当前位置：首页 > news >正文

查找重复代码[A卷-hw_od]

news 来源：原创 2025/4/24 12:00:12

题目描述

小明负责维护项目下的代码，需要查找出重复代码，用以支撑后续的代码优化，请你帮助小明找出重复的代码。重复代码查找方法：以字符串形式给定两行代码（字符串长度 1 < length <= 100，由英文字母、数字和空格组成），找出两行代码中的最长公共子串。注：如果不存在公共子串，返回空字符串

输入描述

输入的参数text1, text2分别表示两行代码

输出描述

输出任一最长公共子串

用例1

输入:

hello123world
hello123abc4

输出:

hello123

用例2

输入:

private_void_method
public_void_method

输出:

_void_method

用例3

输入:

hiworld
hiweb

输出:

hiw

代码如下(C++版):

#include <bits/stdc++.h>
using namespace std;

string solve(const string &s1, const string &s2) {
    int m = s1.size(), n = s2.size();
    vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
    int maxLen = 0, endIdx = 0; //维护最大长度和终止索引

    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (s1[i - 1] == s2[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1] + 1;
                if (dp[i][j] > maxLen) {
                    maxLen = dp[i][j];
                    endIdx = j;
                }
            }
        }
    }
    return (maxLen > 0) ? s1.substr(endIdx - maxLen, maxLen) : "";
}

int main() {
    string s1, s2;
    getline(cin, s1);
    getline(cin, s2);
    string res= solve(s1, s2);
    cout << res << endl;
    return 0;
}

记忆化搜素(C++):

#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> memo;
int maxLen = 0, endIdx = 0;
int dfs(const string &s1, const string &s2, int i, int j) {
    if (i < 0 || j < 0) return 0;
    if (memo[i][j] != -1) return memo[i][j];
    
    if (s1[i] == s2[j]) {
        memo[i][j] = dfs(s1, s2, i - 1, j - 1) + 1;
        if (memo[i][j] > maxLen) {
            maxLen = memo[i][j];
            endIdx = i + 1;
        }
    } else {
        memo[i][j] = 0;
    }
    return memo[i][j];
}

string solve(const string &s1, const string &s2) {
    int m = s1.size(), n = s2.size();
    memo.assign(m, vector<int>(n, -1));
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            dfs(s1, s2, i, j);
        }
    }
    
    return (maxLen > 0) ? s1.substr(endIdx - maxLen, maxLen) : "";
}

int main() {
    string s1, s2;
    getline(cin, s1);
    getline(cin, s2);
    string res= solve(s1, s2);
    cout << res << endl;
    return 0;
}

JAVA版:

import java.util.*;
class solve {
    public String LongestPublicSubstr(String s1, String s2) {
        int m = s1.length(), n = s2.length();
        int[][] dp = new int[m + 1][n + 1];
        int maxLen = 0, endIdx = 0;
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    if (dp[i][j] > maxLen) {
                        maxLen = dp[i][j];
                        endIdx = i;
                    }
                }
            }
        }
        return (maxLen > 0) ? s1.substring(endIdx - maxLen, endIdx) : "";
    }
}
public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s1 = sc.nextLine();
        String s2 = sc.nextLine();
        solve res=new solve();
        System.out.println(res.LongestPublicSubstr(s1, s2));
    }
}

记忆化搜素(JAVA):

import java.util.*;

public class Main {
    private static int[][] memo;
    private static int maxLen = 0, endIdx = 0;
    public static int dfs(String s1, String s2, int i, int j) {
        if (i < 0 || j < 0) return 0;
        if (memo[i][j] != -1) return memo[i][j];

        if (s1.charAt(i) == s2.charAt(j)) {
            memo[i][j] = dfs(s1, s2, i - 1, j - 1) + 1;
            if (memo[i][j] > maxLen) {
                maxLen = memo[i][j];
                endIdx = i + 1;
            }
        } else {
            memo[i][j] = 0;
        }
        return memo[i][j];
    }

    public static String solve(String s1, String s2) {
        int m = s1.length(), n = s2.length();
        memo = new int[m][n];
        for (int[] row : memo) Arrays.fill(row, -1);
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                dfs(s1, s2, i, j);
            }
        }
        
        return (maxLen > 0) ? s1.substring(endIdx - maxLen, endIdx) : "";
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String s1 = scanner.nextLine();
        String s2 = scanner.nextLine();
        System.out.println(solve(s1, s2));
    }
}

练习链接:1143. 最长公共子序列 - 力扣（LeetCode）

1.上面提供的是LeetCode的练习题，都属于线性DP的类型

2. 本题思路: 上面提供了记忆化搜索和递推的写法

状态定义：dp[i][j]表示以i,j的结尾的最长子串

考虑边界: dp[0][0]=0, dp[m][n]=答案

状态转移: 如果当前s[i]==s[j] , dp[i][j]=dp[i-1][j-1]+1 否则dp[i][j]=0

维护数据: 维护一个结尾索引i/j

注：如果使用的记忆化的解法,memo数组一定不要初始化成0,建议初始化成-1.

如果题目很难, 建议使用记忆化搜索, 但是常见的背包/线性DP之类的题目, 建议直接使用递推