LeetCode100题

LeetCode100

两数之和

遍历数组，以哈希表存数与下标，边存边查，速找和为目标值的两数下标

class Solution {public int[] twoSum(int[] nums, int target) {int[] ans=new int[2];HashMap<Integer,Integer> map=new HashMap<>();for(int i=0;i<nums.length;i++){if(map.containsKey(target-nums[i])){ans[0]=map.get(target-nums[i]);ans[1]=i;return ans;}map.put(nums[i],i);}return ans;}
}

字母异位词分组

class Solution {public List<List<String>> groupAnagrams(String[] strs) {List<List<String>> ret=new ArrayList<>();HashMap<String,List<String>> map=new HashMap<>();for(String a:strs){char[] sc=a.toCharArray();Arrays.sort(sc);String key=String.valueOf(sc);List<String> list=map.getOrDefault(key,new ArrayList());list.add(a);map.put(key,list);}return new ArrayList<List<String>>(map.values());}
}

最长连续序列

class Solution {public int longestConsecutive(int[] nums) {int count=0;HashSet<Integer> set=new HashSet<>();for(int x:nums){set.add(x);}for(int x:set){if(!set.contains(x-1)){int flag=1;while(set.contains(++x)){flag++;}count=Math.max(flag,count);}}return count;}
}

移动零

class Solution {public void moveZeroes(int[] nums) {int slow=0;for(int i=0;i<nums.length;i++){if(nums[i]!=0){nums[slow++]=nums[i];}}Arrays.fill(nums,slow,nums.length,0);}
}

盛最多水的容器

class Solution {public int maxArea(int[] height) {int left=0,right=height.length-1;int ans=0;for(int i=0;i<height.length;i++){int high=right-left;int h=Math.min(height[right],height[left]);ans=Math.max(high*h,ans);if(height[right]<height[left]){right--;}else{left++;}}return ans;}
}

三数之和

class Solution {public List<List<Integer>> threeSum(int[] nums) {List<List<Integer>> ans=new ArrayList<>();if(nums.length<3)return ans;Arrays.sort(nums);for(int i=0;i<nums.length-2;i++){if(nums[i]>0)return ans;if(i>0&&nums[i]==nums[i-1])continue;int left=i+1,right=nums.length-1;while(left<right){int sum=nums[i]+nums[left]+nums[right];if(sum==0){ans.add(new ArrayList(Arrays.asList(nums[i],nums[left],nums[right])));}if(sum>0){while(left<right&&nums[right]==nums[--right]);}else{while(left<right&&nums[left]==nums[++left]);}}}return ans;    }
}

接雨水

class Solution {public int trap(int[] height) {int ans = 0;int[] maxleft = new int[height.length];int[] maxright = new int[height.length];for (int i = 1; i < height.length; i++) {maxleft[i] = Math.max(height[i - 1], maxleft[i - 1]);}for (int i = height.length - 2; i >= 0; i--) {maxright[i] = Math.max(maxright[i + 1], height[i + 1]);}for (int i = 1; i < height.length; i++) {int h = Math.min(maxleft[i], maxright[i]);if (h > height[i]) {ans += h - height[i];}}return ans;}
}

无重复字符的最长子串

class Solution {public int lengthOfLongestSubstring(String s) {int count=0,left=0;HashMap<Character,Integer> map=new HashMap<>();for(int i=0;i<s.length();i++){if(map.containsKey(s.charAt(i))){left=Math.max(left,map.get(s.charAt(i))+1);}count=Math.max(count,i-left+1);map.put(s.charAt(i),i);}return count;}
}

找到字符串中所有字母异位词

class Solution {public List<Integer> findAnagrams(String s, String p) {int len = p.length();List<Integer> ans = new ArrayList<>();char[] pc = p.toCharArray();Arrays.sort(pc);String key = String.valueOf(pc);for (int i = 0; i <= s.length() - len; i++) {char[] sc = s.substring(i, i + len).toCharArray();Arrays.sort(sc);if (key.equals(String.valueOf(sc))) {ans.add(i);}}return ans;}
}

和为 K 的子数组

class Solution {public int subarraySum(int[] nums, int k) {int sum=0,ans=0;HashMap<Integer,Integer> map=new HashMap<>();map.put(0,1);for(int i=0;i<nums.length;i++){sum+=nums[i];if(map.containsKey(sum-k)){ans+=map.get(sum-k);}map.put(sum,map.getOrDefault(sum,0)+1);}return ans;}
}

滑动窗口最大值

class Solution {public int[] maxSlidingWindow(int[] nums, int k) {if(nums.length==0||k==0)return new int[0];int[] res=new int[nums.length-k+1];Deque<Integer> deque=new LinkedList<>();for(int i=0,j=1-k;i<nums.length;i++,j++){if(j>0&&deque.peekFirst()==nums[j-1]){deque.removeFirst();}while(!deque.isEmpty()&&deque.peekLast()<nums[i]){deque.removeLast();}deque.addLast(nums[i]);if(j>=0){res[j]=deque.peekFirst();}}return res;}
}

最大子数组和

class Solution {public int maxSubArray(int[] nums) {int pre=0,ans=Integer.MIN_VALUE;for(int x:nums){pre=Math.max(pre+x,x);ans=Math.max(ans,pre);}return ans;}
}

合并区间

class Solution {public int[][] merge(int[][] intervals) {List<int[]> ans=new ArrayList<>();Arrays.sort(intervals,(a,b)->{return a[0]-b[0];});for(int i=0;i<intervals.length;i++){if(i<intervals.length-1&& intervals[i][1]>=intervals[i+1][0]){intervals[i+1][0]=intervals[i][0];intervals[i+1][1]=Math.max(intervals[i][1],intervals[i+1][1]);}else{ans.add(intervals[i]);}}return ans.toArray(new int[0][]);}
}

轮转数组

class Solution {public void rotate(int[] nums, int k) {int count=nums.length-k%nums.length-1;reverse(nums,0,count);reverse(nums,count+1,nums.length-1);reverse(nums,0,nums.length-1);   }void reverse(int[] nums,int left,int right){while(left<right){int temp=nums[right];nums[right]=nums[left];nums[left]=temp;left++;right--;}}
}

除自身以外数组的乘积

class Solution {public int[] productExceptSelf(int[] nums) {int len=nums.length;int[] ans=new int[len];int[] L=new int[len];int[] R=new int[len];L[0]=1;for(int i=1;i<len;i++){L[i]=L[i-1]*nums[i-1];}R[len-1]=1;for(int i=len-2;i>=0;i--){R[i]=nums[i+1]*R[i+1];}for(int i=0;i<len;i++){ans[i]=L[i]*R[i];}return ans;}
}

缺失的第一个正数

class Solution {public int firstMissingPositive(int[] nums) {// 处理数组为空的情况if (nums == null || nums.length == 0) {return 1;}// 第一次遍历，将元素放到正确的位置for (int i = 0; i < nums.length; i++) {while (nums[i] > 0 && nums[i] <= nums.length && nums[nums[i] - 1] != nums[i]) {int temp = nums[nums[i] - 1];nums[nums[i] - 1] = nums[i];nums[i] = temp;}}// 第二次遍历，检查元素是否在正确位置for (int i = 1; i <= nums.length; i++) {if (nums[i - 1] != i) {return i;}}// 如果前面都没有找到缺失的正整数，那么缺失的就是数组长度加 1return nums.length + 1;}
}

矩阵置零

class Solution {public void setZeroes(int[][] matrix) {boolean[] row=new boolean[matrix.length];boolean[] col=new boolean[matrix[0].length];for(int i=0;i<matrix.length;i++){for(int j=0;j<matrix[0].length;j++){if(matrix[i][j]==0){row[i]=true;col[j]=true;}}}for(int i=0;i<matrix.length;i++){for(int j=0;j<matrix[0].length;j++){if(row[i]||col[j]){matrix[i][j]=0;}}}}
}

螺旋矩阵

class Solution {public List<Integer> spiralOrder(int[][] matrix) {List<Integer> ans = new ArrayList<>();int up = 0, down = matrix.length - 1, left = 0, right = matrix[0].length - 1;while (true) {for (int i = left; i <= right; i++) {ans.add(matrix[up][i]);}up++;if (up > down)break;for (int i = up; i <= down; i++) {ans.add(matrix[i][right]);}right--;if (right < left)break;for (int i = right; i >= left; i--) {ans.add(matrix[down][i]);}down--;if (down < up)break;for (int i = down; i >= up; i--) {ans.add(matrix[i][left]);}left++;if (left > right)break;}return ans;}
}

旋转图像

class Solution {public void rotate(int[][] matrix) {int n=matrix.length-1;int[][] temp=new int[matrix.length][matrix.length];for(int i=0;i<matrix.length;i++){temp[i]=matrix[i].clone();}for(int i=0;i<matrix.length;i++){for(int j=0;j<matrix[0].length;j++){matrix[j][n-i]=temp[i][j];}}}
}

搜索二维矩阵 II

class Solution {public boolean searchMatrix(int[][] matrix, int target) {int i=0,j=matrix[0].length-1;while(j>=0&&i<matrix.length){if(target==matrix[i][j]){return true;}else if(target>matrix[i][j]){i++;}else{j--;}}return false;}
}

链表部分：

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/

相交链表

public class Solution {public ListNode getIntersectionNode(ListNode headA, ListNode headB) {ListNode l1=headA;ListNode l2=headB;while(l1!=l2){l1=l1==null?headB:l1.next;l2=l2==null?headA:l2.next;}return l1;}
}

反转链表

class Solution {public ListNode reverseList(ListNode head) {ListNode cur=head,p=null;while(cur!=null){ListNode temp=cur.next;cur.next=p;p=cur;cur=temp;}return p;   }
}

回文链表

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public boolean isPalindrome(ListNode head) {if (head == null || head.next == null) {return true;}//快慢指针找到中间节点ListNode fast = head, slow = head;while (fast.next != null && fast.next.next != null) {fast = fast.next.next;slow = slow.next;}// 翻转后半部分链表ListNode prev = null;ListNode cur = slow.next;while (cur != null) {ListNode temp = cur.next;cur.next = prev;prev = cur;cur = temp;}//分别从链表开头与翻转后的中间节点开始向后比较ListNode p1 = head;ListNode p2 = prev;while (p2 != null) {if (p1.val != p2.val) {return false;}p1 = p1.next;p2 = p2.next;}return true;}
}

环形链表

public class Solution {public boolean hasCycle(ListNode head) {if (head == null || head.next == null) {return false;}ListNode slow = head;ListNode fast = head.next;while (slow != fast) {if (fast == null || fast.next == null) {return false;}slow = slow.next;fast = fast.next.next;}return true;}
}

环形链表 II

public class Solution {public ListNode detectCycle(ListNode head) {Set<ListNode> set=new HashSet<>();ListNode cur=head;while(cur!=null){if(set.contains(cur)){return cur;}else{set.add(cur);}cur=cur.next;}return null;}
}

合并两个有序链表

class Solution {public ListNode mergeTwoLists(ListNode list1, ListNode list2) {ListNode p=list1,q=list2;ListNode prehead = new ListNode(-1);ListNode prev = prehead;while(p!=null&&q!=null){if(p.val<=q.val){prev.next=p;p=p.next;}else{prev.next=q;q=q.next;}prev=prev.next;}prev.next = p == null ? q : p;return prehead.next;}
}

两数相加

class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {ListNode pre = new ListNode(0);ListNode cur = pre;int carry=0;while(l1!=null||l2!=null){int a=l1==null?0:l1.val;int b=l2==null?0:l2.val;cur.next=new ListNode((a+b+carry)%10);carry=(a+b+carry)/10;if (l1 != null) l1 = l1.next;if (l2 != null) l2 = l2.next;// 移动到结果链表的下一个节点cur = cur.next;}if(carry==1)cur.next=new ListNode(1);return pre.next;}
}

删除链表的倒数第 N 个结点

class Solution {public ListNode removeNthFromEnd(ListNode head, int n) {ListNode dump=new ListNode(0);dump.next=head;ListNode slow=dump,fast=dump;while(n>0){fast=fast.next;n--;}while(fast.next!=null){fast=fast.next;slow=slow.next;}slow.next=slow.next.next;return dump.next;}
}

两两交换链表中的节点

class Solution {public ListNode swapPairs(ListNode head) {ListNode node=new ListNode(-1,head);ListNode pre=node;while(pre.next!=null&&pre.next.next!=null){ListNode l1=pre.next,l2=pre.next.next;l1.next=l2.next;l2.next=l1;pre.next=l2;pre=l1;}return node.next;}
}

K 个一组翻转链表

class Solution {public ListNode reverseKGroup(ListNode head, int k) {if (head == null || k <= 1)return head;ListNode dummy = new ListNode(0);dummy.next = head;ListNode prevGroupTail = dummy;int n = countNodes(head);while (n >= k) {ListNode groupHead = prevGroupTail.next;ListNode cur = groupHead;ListNode pre = null;for (int i = 0; i < k; i++) {ListNode temp = cur.next;cur.next = pre;pre = cur;cur = temp;}prevGroupTail.next = pre;groupHead.next = cur;prevGroupTail = groupHead;n -= k;}return dummy.next;}private int countNodes(ListNode head) {int count = 0;// 遍历链表并计算节点数while (head != null) {count++;head = head.next;}return count;}
}

138. 随机链表的复制

class Solution {public Node copyRandomList(Node head) {HashMap<Node,Node> map=new HashMap<>();Node cur=head;while(cur!=null){map.put(cur,new Node(cur.val));cur=cur.next;}cur=head;while(cur!=null){map.get(cur).next=map.get(cur.next);map.get(cur).random=map.get(cur.random);cur=cur.next;}return map.get(head);}
}

148. 排序链表

class Solution {public ListNode sortList(ListNode head) {if(head==null||head.next==null)return head;ListNode slow=head;ListNode fast=head;while(fast.next!=null&&fast.next.next!=null){slow=slow.next;fast=fast.next.next;}ListNode nhead=slow.next;slow.next=null;ListNode left= sortList(head);ListNode right=sortList(nhead);ListNode dummy=new ListNode(-1);ListNode cur=dummy;while(left!=null&&right!=null){if(left.val<right.val){cur.next=left;left=left.next;}else{cur.next=right;right=right.next;}cur=cur.next;}cur.next=left==null?right:left;return dummy.next;}
}

合并 K 个升序链表

class Solution {public ListNode mergeKLists(ListNode[] lists) {if(lists==null||lists.length==0)return null;PriorityQueue<ListNode> queue=new PriorityQueue<>(lists.length,Comparator.comparingInt(o->o.val));for(ListNode node:lists){if(node!=null)queue.add(node);}ListNode pre=new ListNode(-1);ListNode p=pre;while(!queue.isEmpty()){p.next=queue.poll();p=p.next;if(p.next!=null)queue.add(p.next);}return pre.next;}
}

146. LRU 缓存

class LRUCache {HashMap<Integer,Node> map=new LinkedHashMap<>();int _capacity;Node head,tail;class Node{int key;int values;Node pre;Node next;Node(int _key,int _values){key=_key;values=_values;}}public LRUCache(int capacity) {_capacity=capacity;head=new Node(-1,-1);tail=new Node(-1,-1);head.next=tail;tail.pre=head;}public int get(int key) {if(map.containsKey(key)){Node node=map.get(key);cutNode(node);insert(node);return node.values;}else{return -1;}}public void put(int key, int value) {if(map.containsKey(key)){Node node=map.get(key);node.values=value;cutNode(node);insert(node);}else{Node newNode=new Node(key,value);map.put(key,newNode);insert(newNode);if(map.size()>_capacity){Node node=cuttail();map.remove(node.key);}}}void cutNode(Node node){node.pre.next=node.next;node.next.pre=node.pre;node.pre=null;node.next=null;}void insert(Node node){node.next=head.next;head.next.pre=node;head.next=node;node.pre=head;}Node cuttail(){Node last=tail.pre;last.pre.next=tail;tail.pre=last.pre;last.pre=null;last.next=null;return last;}
}

树的定义：

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/

二叉树的中序遍历

class Solution {public List<Integer> inorderTraversal(TreeNode root) {List<Integer> res = new ArrayList<Integer>();inorder(root, res);return res;}public void inorder(TreeNode root, List<Integer> res) {if (root == null) {return;}inorder(root.left, res);res.add(root.val);inorder(root.right, res);}
}

104. 二叉树的最大深度

class Solution {public int maxDepth(TreeNode root) {if(root==null){return 0;}int maxleft=maxDepth(root.left);int maxright=maxDepth(root.right);return Math.max(maxleft,maxright)+1;}
}

226. 翻转二叉树

class Solution {public TreeNode invertTree(TreeNode root) {if(root==null)return null;TreeNode left=invertTree(root.left);TreeNode right=invertTree(root.right);root.left=right;root.right=left;return root;}
}

101. 对称二叉树

class Solution {public boolean isSymmetric(TreeNode root) {return check(root, root);}public boolean check(TreeNode p, TreeNode q) {if (p == null && q == null) {return true;}if ((p == null || q == null)||(p.val != q.val)) {return false;}return  check(p.left, q.right) && check(p.right, q.left);}
}

543. 二叉树的直径

class Solution {int ans;public int diameterOfBinaryTree(TreeNode root) {ans = 1;depth(root);return ans - 1;}public int depth(TreeNode node) {if (node == null) {return 0; // 访问到空节点了，返回0}int L = depth(node.left); // 左儿子为根的子树的深度int R = depth(node.right); // 右儿子为根的子树的深度ans = Math.max(ans, L+R+1); // 计算d_node即L+R+1 并更新ansreturn Math.max(L, R) + 1; // 返回该节点为根的子树的深度}
}

102. 二叉树的层序遍历

class Solution {public List<List<Integer>> levelOrder(TreeNode root) {List<List<Integer>> res=new ArrayList<>();Queue<TreeNode> queue=new LinkedList<>();if(root!=null)queue.offer(root);while(!queue.isEmpty()){List<Integer> tem=new ArrayList<>();for(int qsize=queue.size();qsize>0;qsize--){TreeNode node=queue.poll();if(node.left!=null)queue.offer(node.left);if(node.right!=null)queue.offer(node.right);tem.add(node.val);}res.add(tem);}return res;}
}

108. 将有序数组转换为二叉搜索树

class Solution {public TreeNode sortedArrayToBST(int[] nums) {return helper(nums, 0, nums.length - 1);}public TreeNode helper(int[] nums, int left, int right) {if (left > right) {return null;}// 总是选择中间位置左边的数字作为根节点int mid = (left + right) / 2;TreeNode root = new TreeNode(nums[mid]);root.left = helper(nums, left, mid - 1);root.right = helper(nums, mid + 1, right);return root;}
}

98. 验证二叉搜索树

class Solution {public boolean isValidBST(TreeNode root) {return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);}public boolean isValidBST(TreeNode node, long lower, long upper) {if (node == null) {return true;}if (node.val <= lower || node.val >= upper) {return false;}return isValidBST(node.left, lower, node.val) && isValidBST(node.right, node.val, upper);}
}

230. 二叉搜索树中第 K 小的元素

class Solution {int count, ans;public int kthSmallest(TreeNode root, int k) {count = k;inOrder(root);return ans;}void inOrder(TreeNode root) {if (root == null || count == 0)return;inOrder(root.left);count--;if (count == 0) {ans = root.val;return;}inOrder(root.right);}
}

199. 二叉树的右视图

class Solution {List<Integer> ans = new ArrayList<>();public List<Integer> rightSideView(TreeNode root) {dfs(root, 0);return ans;}private void dfs(TreeNode root, int depth) {if (root == null) return;if (depth == ans.size()) ans.add(root.val);dfs(root.right, depth + 1);dfs(root.left, depth + 1);}
}

114. 二叉树展开为链表

class Solution {public void flatten(TreeNode root) {if (root == null) {return;}Stack<TreeNode> s = new Stack<TreeNode>();s.push(root);TreeNode pre = null;while (!s.isEmpty()) {TreeNode temp = s.pop();// 将指针的柚子树指向当前节点if (pre != null) {pre.right = temp;pre.left = null;}if (temp.right != null) {s.push(temp.right);}if (temp.left != null) {s.push(temp.left);}// 将指针移动到当前位置pre = temp;}}
}

105. 从前序与中序遍历序列构造二叉树

class Solution {Map<Integer, Integer> map = new HashMap<>();public TreeNode buildTree(int[] preorder, int[] inorder) {for(int i=0;i<inorder.length;i++){map.put(inorder[i],i);}return buildTwo(preorder,0,preorder.length-1,0,inorder.length-1);}TreeNode buildTwo(int[] preorder,int preorder_left,int preorder_right,int inorder_left,int inorder_right){if(preorder_left>preorder_right)return null;TreeNode root=new TreeNode(preorder[preorder_left]);int pindex=map.get(preorder[preorder_left]);root.left=buildTwo(preorder,preorder_left+1,pindex-1-inorder_left+preorder_left+1,inorder_left,pindex-1);root.right=buildTwo(preorder,pindex-1-inorder_left+preorder_left+1+1,preorder_right,pindex+1,inorder_right);return root;}
}

437. 路径总和 III

class Solution {int ans=0;HashMap<Long,Integer> map=new HashMap<>();public int pathSum(TreeNode root, int targetSum) {map.put(0L,1);dfs(root,targetSum,0L);return ans;}void dfs(TreeNode root,int targetSum,Long sum){if(root==null)return ;sum+=root.val;ans+=map.getOrDefault(sum-targetSum,0);map.put(sum,map.getOrDefault(sum,0)+1);dfs(root.left,targetSum,sum);dfs(root.right,targetSum,sum);map.put(sum,map.getOrDefault(sum,0)-1);return;}
}

236. 二叉树的最近公共祖先

class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if(root==null||root==p||root==q)return root;TreeNode left=lowestCommonAncestor(root.left,p,q);TreeNode right=lowestCommonAncestor(root.right,p,q);if(left==null&&right==null){return null;}if(left==null){return right;}if(right==null){return left;}return root;}
}

124. 二叉树中的最大路径和

class Solution {int maxSum = Integer.MIN_VALUE;public int maxPathSum(TreeNode root) {maxGain(root);return maxSum;}public int maxGain(TreeNode node) {if (node == null) {return 0;}// 递归计算左右子节点的最大贡献值// 只有在最大贡献值大于 0 时，才会选取对应子节点int leftGain = Math.max(maxGain(node.left), 0);int rightGain = Math.max(maxGain(node.right), 0);// 节点的最大路径和取决于该节点的值与该节点的左右子节点的最大贡献值int priceNewpath = node.val + leftGain + rightGain;// 更新答案maxSum = Math.max(maxSum, priceNewpath);// 返回节点的最大贡献值return node.val + Math.max(leftGain, rightGain);}
}

200. 岛屿数量

class Solution {private int res;public int numIslands(char[][] grid) {res = 0;for (int i = 0; i < grid.length; i ++) {for (int j = 0; j < grid[0].length; j ++) {if (grid[i][j] == '1') {dfsGrid(grid, i, j);res ++;}}}return res;}private void dfsGrid(char[][] grid, int row, int col) {if (row<0 || col<0 || row >= grid.length || col >= grid[0].length || grid[row][col] != '1' )return;grid[row][col] = '0';dfsGrid(grid, row - 1, col);dfsGrid(grid, row + 1, col);dfsGrid(grid, row, col - 1);dfsGrid(grid, row, col + 1);}
}

994. 腐烂的橘子

class Solution {int freshCount=0;public int orangesRotting(int[][] grid) {int row=grid.length;int col=grid[0].length;Queue<int[]> queue=new LinkedList<>();for(int i=0;i<row;i++){for(int j=0;j<col;j++){if(grid[i][j]==1){freshCount++;}else if(grid[i][j]==2){queue.add(new int[]{i,j});}}}int round=0;while(freshCount>0&&!queue.isEmpty()){round++;int n=queue.size();for(int i=0;i<n;i++){int[] orange=queue.poll();int x=orange[0];int y=orange[1];if(x-1>=0&&grid[x-1][y]==1){grid[x-1][y]=2;freshCount--;queue.add(new int[]{x-1,y});}if(x+1<row&&grid[x+1][y]==1){grid[x+1][y]=2;freshCount--;queue.add(new int[]{x+1,y});}if(y-1>=0&&grid[x][y-1]==1){grid[x][y-1]=2;freshCount--;queue.add(new int[]{x,y-1});}if(y+1<col&&grid[x][y+1]==1){grid[x][y+1]=2;freshCount--;queue.add(new int[]{x,y+1});}}}if(freshCount>0){return -1;}else{return round;}}
}

207. 课程表

class Solution {public boolean canFinish(int numCourses, int[][] prerequisites) {// 入度数组，用于记录每门课程的入度int[] inDegree=new int[numCourses];// 邻接表，存储每门课程的后续课程List<List<Integer>> adjList=new ArrayList<>();for(int i=0;i<numCourses;i++){adjList.add(new ArrayList<>());}// 计算每门课程的入度，并构建邻接表for(int[] course:prerequisites){inDegree[course[0]]++;adjList.get(course[1]).add(course[0]);}// 存储入度为 0 的课程的队列Queue<Integer> queue=new LinkedList<>();for(int i=0;i<numCourses;i++){if(inDegree[i]==0){queue.offer(i);}}// 记录已完成课程的数量int count=0;while(!queue.isEmpty()){int selectCourse=queue.poll();count++;// 获取当前课程的后续课程列表for(int x:adjList.get(selectCourse)){// 后续课程的入度减 1,将入度为0的加入队列inDegree[x]--;if(inDegree[x]==0){queue.offer(x);}}}if(count==numCourses)return true;return false;}
}

208. 实现 Trie (前缀树)

class Trie {class treeNode{boolean island;treeNode[] child;treeNode(){island=false;child=new treeNode[26];}}treeNode root;public Trie() {root=new treeNode();}public void insert(String word) {treeNode node=root;for(char i:word.toCharArray()){if(node.child[i-'a']==null){node.child[i-'a']=new treeNode();}node=node.child[i-'a'];}node.island=true;}public boolean search(String word) {treeNode node=root;for(char c:word.toCharArray()){if(node.child[c-'a']==null){return false;}node=node.child[c-'a'];}return node.island;}public boolean startsWith(String prefix) {treeNode node=root;for(char c:prefix.toCharArray()){if(node.child[c-'a']==null){return false;}node=node.child[c-'a'];}return true;}
}

46. 全排列

class Solution {List<List<Integer>> res = new LinkedList<>();public List<List<Integer>> permute(int[] nums) {var track = new ArrayList<Integer>();backtrack(nums, track);return res;}private void backtrack(int[] nums, ArrayList<Integer> track) {if(track.size() == nums.length) {res.add(new ArrayList<Integer>(track));return;}int n = nums.length;for(int i = 0; i < n; ++i) {if(track.contains(nums[i]))continue;track.add(nums[i]);backtrack(nums, track);track.removeLast();}}
}

78. 子集

class Solution {private final List<List<Integer>> ans = new ArrayList<>();private final List<Integer> path = new ArrayList<>();private int[] nums;public List<List<Integer>> subsets(int[] nums) {this.nums = nums;dfs(0);return ans;}private void dfs(int i) {if (i == nums.length) { // 子集构造完毕ans.add(new ArrayList<>(path)); // 复制 pathreturn;}// 不选 nums[i]dfs(i + 1);// 选 nums[i]path.add(nums[i]);dfs(i + 1);path.remove(path.size() - 1); // 恢复现场}
}

17. 电话号码的字母组合

class Solution {Map<Character, String> phoneMap = new HashMap<Character, String>() {{put('2', "abc");put('3', "def");put('4', "ghi");put('5', "jkl");put('6', "mno");put('7', "pqrs");put('8', "tuv");put('9', "wxyz");}};List<String> ans = new ArrayList<String>();public List<String> letterCombinations(String digits) {if(digits.length()==0)return ans;breakf(digits,0,new StringBuilder());return ans;}void breakf(String digits,int start,StringBuilder temp){if(start==digits.length()){ans.add(String.valueOf(temp));return;}char c=digits.charAt(start);String value=phoneMap.get(c);for(int i=0;i<value.length();i++){temp.append(value.charAt(i));breakf(digits,start+1,temp);temp.deleteCharAt(temp.length()-1);}}
}

39. 组合总和

class Solution {List<List<Integer>> res=new ArrayList<>();List<Integer> path=new ArrayList<>();public List<List<Integer>> combinationSum(int[] candidates, int target) {dfs(candidates,target,0);return res;}void dfs(int[] candidates, int target,int begin){if(target==0){res.add(new ArrayList(path));return;}if(target<0)return;for(int i=begin;i<candidates.length;i++){path.add(candidates[i]);dfs(candidates,target-candidates[i],i);path.removeLast();}}
}

22. 括号生成

class Solution {StringBuilder s=new StringBuilder();List<String> res=new ArrayList<String>();public List<String> generateParenthesis(int n) {dfs(n,0,0);return res;}public void dfs(int n,int left,int right){if(left==n&&right==n){res.add(String.valueOf(s));}if(left<n){s.append('(');dfs(n,left+1,right);s.deleteCharAt(s.length()-1);}if(right<left){s.append(')');dfs(n,left,right+1);s.deleteCharAt(s.length()-1);}}
}

79. 单词搜索

class Solution {public boolean exist(char[][] board, String word) {if (board.length == 0) {return false;}char[] chars = word.toCharArray();for (int i = 0; i < board.length; i++) {for (int j = 0; j < board[0].length; j++) {if (board[i][j] != chars[0])continue;if (dfs(i, j, board, 0, chars)) {return true;}}}return false;}private boolean dfs(int x, int y, char[][] board, int index, char[] chars) {if (x < 0 || x > board.length - 1 ||y < 0 || y > board[0].length - 1 ||board[x][y] != chars[index]) {return false;}if (index == chars.length - 1) {return true;}board[x][y] = '\0';if (dfs(x + 1, y, board, index + 1, chars) || dfs(x - 1, y, board, index + 1, chars)|| dfs(x, y + 1, board, index + 1, chars) || dfs(x, y - 1, board, index + 1, chars))return true;board[x][y] = chars[index];return false;}
}

131. 分割回文串

class Solution {private final List<List<String>> ans = new ArrayList<>();private final List<String> path = new ArrayList<>();private String s;public List<List<String>> partition(String s) {this.s = s;dfs(0, 0);return ans;}// start 表示当前这段回文子串的开始位置private void dfs(int i, int start) {if (i == s.length()) {ans.add(new ArrayList<>(path)); // 复制 pathreturn;}// 不选 i 和 i+1 之间的逗号（i=n-1 时一定要选）if (i < s.length() - 1) {dfs(i + 1, start);}// 选 i 和 i+1 之间的逗号（把 s[i] 作为子串的最后一个字符）if (isPalindrome(start, i)) {path.add(s.substring(start, i + 1));dfs(i + 1, i + 1); // 下一个子串从 i+1 开始path.remove(path.size() - 1); // 恢复现场}}private boolean isPalindrome(int left, int right) {while (left < right) {if (s.charAt(left++) != s.charAt(right--)) {return false;}}return true;}
}

35. 搜索插入位置

class Solution {public int searchInsert(int[] nums, int target) {int left=0,right=nums.length-1;while(left<=right){int mid=(right-left)/2+left;if(nums[mid]==target)return mid;if(nums[mid]>=target){right=mid-1;}if(nums[mid]<target){left=mid+1;}}return left;}
}

74. 搜索二维矩阵

class Solution {public boolean searchMatrix(int[][] matrix, int target) {int n=matrix[0].length;int left=0,right=matrix.length*n-1;while(left<=right){int mid=(right-left)/2+left;int temp=matrix[mid/n][mid%n];if(temp==target)return true;if(temp>target){right=mid-1;}else{left=mid+1;}}return false;}
}

34. 在排序数组中查找元素的第一个和最后一个位置

class Solution {/*** 二分查找*/public int[] searchRange(int[] nums, int target) {//寻找左边界(这里寻找第一个 >= target的索引)int leftIndex = search(nums, target);if (leftIndex >= nums.length || nums[leftIndex] != target){return new int[]{-1, -1};}//寻找右边界(这里寻找第一个 >= target+1的索引)int rightIndex = search(nums, target + 1);return new int[]{leftIndex, rightIndex - 1};}/*** 寻找第一个>=目标值的索引, 找不到则返回数组长度*/private int search(int[] nums, int target) {int left = 0;int right = nums.length - 1;while (left <= right){int mid = (right - left) / 2 + left;if (nums[mid] >= target){right = mid - 1;} else {left = mid + 1;}}return left;}
}

33. 搜索旋转排序数组

class Solution {public int search(int[] nums, int target) {int len = nums.length;if(len == 0) return -1;int left = 0, right = len - 1;// 1. 首先明白，旋转数组后，从中间划分，一定有一边是有序的。// 2. 由于一定有一边是有序的，所以根据有序的两个边界值来判断目标值在有序一边还是无序一边// 3. 这题找目标值，遇到目标值即返回// 4. 注意：由于有序的一边的边界值可能等于目标值，所以判断目标值是否在有序的那边时应该加个等号(在二分查找某个具体值得时候如果把握不好边界值，可以再每次查找前判断下边界值，也就是while循环里面的两个if注释)while(left <= right){// if(nums[left] == target) return left;// if(nums[right] == target) return right;int mid = left + (right - left) / 2;if(nums[mid] == target) return mid;// 右边有序if(nums[mid] < nums[right]){// 目标值在右边if(target > nums[mid] && target <= nums[right]){left = mid + 1;// 目标值在左边}else{right = mid - 1;}// 左边有序}else{// 目标值在左边if(target >= nums[left] && target < nums[mid]){right = mid - 1;// 目标值在右边}else{left = mid + 1;}}}return -1;}
}

153. 寻找旋转排序数组中的最小值

class Solution {public int findMin(int[] nums) {int ans=9999;int left=0,right=nums.length-1;while(left<=right){int mid=(right-left)/2+left;if(nums[mid]<nums[right]){right=mid-1;ans=Math.min(ans,nums[mid]);}else{   ans=Math.min(ans,nums[left]);left=mid+1;}}return ans;}
}

4. 寻找两个正序数组的中位数

class Solution {public double findMedianSortedArrays(int[] nums1, int[] nums2) {int m = nums1.length;int n = nums2.length;// 确保 nums1 是较短的数组，这样可以减少二分查找的范围if (m > n) {return findMedianSortedArrays(nums2, nums1);}int left = 0, right = m;while (left <= right) {// 在 nums1 中划分的位置int partition1 = (left + right) / 2;// 在 nums2 中划分的位置，使得左右两部分元素数量大致相等int partition2 = (m + n + 1) / 2 - partition1;// 获取划分位置左右两侧的元素值int maxLeft1 = (partition1 == 0) ? Integer.MIN_VALUE : nums1[partition1 - 1];int minRight1 = (partition1 == m) ? Integer.MAX_VALUE : nums1[partition1];int maxLeft2 = (partition2 == 0) ? Integer.MIN_VALUE : nums2[partition2 - 1];int minRight2 = (partition2 == n) ? Integer.MAX_VALUE : nums2[partition2];// 检查划分是否正确if (maxLeft1 <= minRight2 && maxLeft2 <= minRight1) {// 如果总元素个数是奇数，中位数是左半部分的最大值if ((m + n) % 2 == 1) {return Math.max(maxLeft1, maxLeft2);} else {// 如果总元素个数是偶数，中位数是左半部分最大值和右半部分最小值的平均值return (Math.max(maxLeft1, maxLeft2) + Math.min(minRight1, minRight2)) / 2.0;}} else if (minRight1 > maxLeft2) {// 划分位置在 nums1 中太靠右，需要向左移动right = partition1 - 1;} else {// 划分位置在 nums1 中太靠左，需要向右移动left = partition1 + 1;}}return -1;}
}

20. 有效的括号

class Solution {public boolean isValid(String s) {if(s.isEmpty())return true;Stack<Character> stack=new Stack<Character>();for(char c:s.toCharArray()){if(c=='('){stack.push(')');}else if(c=='{'){stack.push('}');}else if(c=='['){stack.push(']');}else if(stack.isEmpty()||stack.pop()!=c){return false;}}return true;}
}

155. 最小栈

class MinStack {Stack<Integer> s1=new Stack<>();Stack<Integer> s2=new Stack<>();public MinStack() {}public void push(int val) {s1.push(val);if(s2.isEmpty()||s2.peek()>=val){s2.push(val);}}public void pop() {if (s1.pop().equals(s2.peek())) {s2.pop();}}public int top() {return s1.peek();}public int getMin() {return s2.peek();}
}

739. 每日温度

class Solution {public int[] dailyTemperatures(int[] temperatures) {Stack<Integer> st=new Stack<>();int[] ans=new int[temperatures.length];for(int i=0;i<temperatures.length;i++){while(!st.isEmpty()&&temperatures[st.peek()] <temperatures[i]){int last=st.peek();ans[st.pop()]=i-last;}st.push(i);}return ans;}
}

215. 数组中的第K个最大元素

class Solution {public int findKthLargest(int[] nums, int k) {PriorityQueue<Integer> minHeap = new PriorityQueue<>(k);for (int num : nums) {if (minHeap.size() < k) {minHeap.offer(num);} else if (num > minHeap.peek()) {minHeap.poll();minHeap.offer(num);}}return minHeap.peek();}
}

347. 前 K 个高频元素

class Solution {public int[] topKFrequent(int[] nums, int k) {HashMap<Integer,Integer> map=new HashMap<>();for(int i=0;i<nums.length;i++){map.put(nums[i],map.getOrDefault(nums[i],0)+1);}PriorityQueue<Integer> queue=new PriorityQueue<>(k,(o1,o2)->(map.get(o2)-map.get(o1)));for(int x:map.keySet()){queue.add(x);}int[] ans=new int[k];for(int i=0;i<k;i++){ans[i]=queue.poll();}return ans;}
}

121. 买卖股票的最佳时机

class Solution {public int maxProfit(int[] prices) {int cost=prices[0],profit=0;for(int price:prices){cost=Math.min(cost,price);profit=Math.max(profit,price-cost);}return profit;}
}

55. 跳跃游戏

class Solution {public boolean canJump(int[] nums) {int start=0;int end=1;int count=0;while(end<nums.length){int maxpos=0;for(int i=start;i<end;i++){maxpos=Math.max(maxpos,nums[i]+i);}if(maxpos==0)return false;start=end;end=maxpos+1; }return true;}
}

45. 跳跃游戏 II

class Solution {public int jump(int[] nums) {int start=0;int end=1;int count=0;while(end<nums.length){int maxpos=0;for(int i=start;i<end;i++){maxpos=Math.max(nums[i]+i,maxpos);}start=end;end=1+maxpos;count++;}return count;}
}

763. 划分字母区间

class Solution {public List<Integer> partitionLabels(String s) {int[] last = new int[26];for (int i = 0; i < s.length(); i++) {last[s.charAt(i) - 'a'] = i;}List<Integer> ans = new ArrayList<>();int maxpoi = 0, start = 0;for (int i = 0; i < s.length(); i++) {maxpoi = Math.max(maxpoi, last[s.charAt(i) - 'a']);if (i == maxpoi) {ans.add(maxpoi - start + 1);start = i + 1;}}return ans;}
}

70. 爬楼梯

class Solution {//当楼梯为1时，那就是1；当楼梯为2时，那就是2；//当楼梯为3时，那就是1+2=3；当楼梯为4时，那就是2+3=5。当楼梯为5时，那就是3+5等于8.public int climbStairs(int n) {int a=1,b=1,sum=0;for(int i=1;i<n;i++){sum=a+b;a=b;b=sum;}return b;}
}

118. 杨辉三角

class Solution {public List<List<Integer>> generate(int numRows) {// 创建一个存储所有行的列表List<List<Integer>> ans = new ArrayList<>();// 外层循环遍历每一行for (int i = 0; i < numRows; i++) {// 创建当前行的临时列表List<Integer> temp = new ArrayList<>();// 每行的第一个元素总是1temp.add(1);// 内层循环计算当前行的中间元素for (int j = 1; j < i; j++) {// 当前元素是上一行相邻两个元素的和temp.add(ans.get(i - 1).get(j - 1) + ans.get(i - 1).get(j));}// 如果当前行不是第一行，添加最后一个元素1if (i > 0) temp.add(1);// 将当前行添加到结果列表中ans.add(temp);}// 返回生成的杨辉三角return ans;}
}

198. 打家劫舍

class Solution {public int rob(int[] nums) {if (nums.length == 0) {return 0;}// 子问题：// f(k) = 偷 [0..k) 房间中的最大金额// f(0) = 0// f(1) = nums[0]// f(k) = max{ rob(k-1), nums[k-1] + rob(k-2) }int N = nums.length;int[] dp = new int[N + 1];dp[0] = 0;dp[1] = nums[0];for (int k = 2; k <= N; k++) {dp[k] = Math.max(dp[k - 1], nums[k - 1] + dp[k - 2]);}return dp[N];}
}

322. 零钱兑换

class Solution {public int coinChange(int[] coins, int amount) {int[] dp = new int[amount + 1];// 初始化数组Arrays.fill(dp, -1);dp[0] = 0;for (int i = 1; i <= amount; i++) {for (int coin : coins) {if (i >= coin && dp[i - coin] != -1) {if (dp[i] == -1 || dp[i] > dp[i - coin] + 1) {dp[i] = dp[i - coin] + 1;}}}}return dp[amount];}
}

139. 单词拆分

class Solution {public boolean wordBreak(String s, List<String> wordDict) {HashSet<String> wordDictSet = new HashSet(wordDict);boolean[] dp = new boolean[s.length() + 1];dp[0] = true;for (int i = 0; i <= s.length(); i++) {for (int j = 0; j < i; j++) {if (dp[j] && wordDictSet.contains(s.substring(j, i))) {dp[i] = true;break;}}}return dp[s.length()];}
}

300. 最长递增子序列

class Solution {public int lengthOfLIS(int[] nums) {/*贪心法加二分查找,如果当前元素比最后一个大，那就直接添加，否则就在前边有序中找适合他的位置，替换那个位置，长度始终不会比当前小，不断替换元素*/int len=nums.length,cur=1;int[] dp=new int[len+1];dp[cur]=nums[0];for(int i=1;i<len;i++){if(dp[cur]<nums[i]){dp[++cur]=nums[i];}else{int l=1,r=cur,pos=0;while(l<=r){int mid=(r-l)/2+l;if(dp[mid]>nums[i]){r=mid-1;}else{pos=mid;l=mid+1;}}dp[pos+1]=nums[i];}}return cur;}
}

152. 乘积最大子数组

class Solution {public int maxProduct(int[] nums) {int max=Integer.MIN_VALUE;int imax=1,imin=1;for(int num:nums){if(num<0){int temp=imax;imax=imin;imin=temp;}imax=Math.max(imax*num,num);imin=Math.min(imin*num,num);max=Math.max(imax,max);}return max;}
}

416. 分割等和子集

class Solution {public boolean canPartition(int[] nums) {int sum=0;for(int x:nums){sum+=x;}if(sum%2==1)return false;sum=sum/2;int[] dp=new int[sum+1];for(int i=0;i<nums.length;i++){for(int j=sum;j>=nums[i];j--){dp[j]=Math.max(dp[j],dp[j-nums[i]]+nums[i]);}}if(dp[sum]==sum)return true;return false;}
}

62. 不同路径

class Solution {public int uniquePaths(int m, int n) {int[][] grid=new int[m][n];for(int i=0;i<m;i++){grid[i][0]=1;}Arrays.fill(grid[0], 1);for(int i=1;i<m;i++){for(int j=1;j<n;j++){grid[i][j]=grid[i-1][j]+grid[i][j-1];}}return grid[m-1][n-1];}
}

64. 最小路径和

class Solution {public int minPathSum(int[][] grid) {for(int i=0;i<grid.length;i++){for(int j=0;j<grid[0].length;j++){if(i==0&&j==0)continue;if(i==0)grid[i][j]=grid[i][j-1]+grid[i][j];if(j==0)grid[i][j]=grid[i][j]+grid[i-1][j];if(i!=0&&j!=0)grid[i][j]=grid[i][j]+Math.min(grid[i-1][j],grid[i][j-1]);}}return grid[grid.length-1][grid[0].length-1];}
}

1143. 最长公共子序列

class Solution {public int longestCommonSubsequence(String text1, String text2) {int m=text1.length(),n=text2.length();int[][] dp=new int[m+1][n+1];Arrays.fill(dp[0],0);for(int i=0;i<m;i++){dp[i][0]=0;}for(int i=1;i<=m;i++){for(int j=1;j<=n;j++){if(text1.charAt(i-1)==text2.charAt(j-1)){dp[i][j]= dp[i-1][j-1]+1;}else{dp[i][j]=Math.max(dp[i-1][j],dp[i][j-1]);}}}return dp[m][n];}
}

72. 编辑距离

class Solution {public int minDistance(String word1, String word2) {int n = word1.length();int m = word2.length();// 有一个字符串为空串if (n * m == 0) {return n + m;}// DP 数组int[][] D = new int[n + 1][m + 1];// 边界状态初始化for (int i = 0; i < n + 1; i++) {D[i][0] = i;}for (int j = 0; j < m + 1; j++) {D[0][j] = j;}// 计算所有 DP 值for (int i = 1; i < n + 1; i++) {for (int j = 1; j < m + 1; j++) {int left = D[i - 1][j] + 1;int down = D[i][j - 1] + 1;int left_down = D[i - 1][j - 1];if (word1.charAt(i - 1) != word2.charAt(j - 1)) {left_down += 1;}D[i][j] = Math.min(left, Math.min(down, left_down));}}return D[n][m];}
}

136. 只出现一次的数字

class Solution {public int singleNumber(int[] nums) {int sigle=0;for(int x:nums){sigle=sigle^x;}return sigle;}
}

169. 多数元素

class Solution {public int majorityElement(int[] nums) {int a=1,key=0;key=nums[0];for(int i=1;i<nums.length;i++){if(key==nums[i]){a++;}else{a--;}if(a<0){key=nums[i];a=1;}}return key;}
}

75. 颜色分类

class Solution {public void sortColors(int[] nums) {int r=0,w=0,b=0;for(int x:nums){if(x==0)r++;if(x==1)w++;if(x==2)b++;}Arrays.fill(nums,0,r,0);Arrays.fill(nums,r,w+r,1);Arrays.fill(nums,r+w,r+w+b,2);}
}

31. 下一个排列

class Solution {// 寻找下一个排列public void nextPermutation(int[] nums) {int pos = -1;// 寻找第一个比后边小的数for (int i = nums.length - 1; i > 0; i--) {if (nums[i] > nums[i - 1]) {pos = i - 1;break;}}// 没找到直接反转数组if (pos == -1) {reverse(nums, 0, nums.length - 1);return;}// 找到 交换他和前边的数，然后翻转pos之后的数for (int i = nums.length - 1; i > 0; i--) {if (nums[pos] < nums[i]) {swap(nums, i, pos);reverse(nums, pos + 1, nums.length - 1);return;}}}void swap(int[] nums, int a, int b) {int temp = nums[a];nums[a] = nums[b];nums[b] = temp;}void reverse(int[] nums, int left, int right) {while (left < right) {int temp = nums[left];nums[left] = nums[right];nums[right] = temp;left++;right--;}}
}

287. 寻找重复数

class Solution {public int findDuplicate(int[] nums) {int slow = nums[0], fast = nums[nums[0]];while (slow != fast) {slow = nums[slow];fast = nums[nums[fast]];}fast = 0;while (slow != fast) {slow = nums[slow];fast = nums[fast];}return fast;}
}