【日常刷题/动态规划C++】交错字符串，两个字符串的最小ASSCLl删除和

一，交错字符串

1，题目解析

判断字符串s3是否可以由两个字符串s1和s2交替拼接而成

 

2，解题思路 

 

3，代码 

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        int m = s1.size(), n = s2.size();
        if (m + n != s3.size())
            return false;
        s1 = " " + s1, s2 = " " + s2, s3 = " " + s3;

        vector<vector<bool>> dp(m + 1, vector<bool>(n + 1));
        dp[0][0] = true;

        // s1为空
        for (int i = 1; i <= n; i++)
            if (s2[i] == s3[i])
                dp[0][i] = true;
            else
                break;

        // s2为空
        for (int j = 1; j <= m; j++)
            if (s1[j] == s3[j])
                dp[j][0] = true;
            else
                break;

        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++) {
                if (s3[i + j] == s2[j] && dp[i][j - 1])
                    dp[i][j] = true;
                if (s3[i + j] == s1[i] && dp[i - 1][j])
                    dp[i][j] = true;
            }

        return dp[m][n];
    }
};

二，两个字符串的最小ASCLL删除和

1,题目解析

2，解题思路

3，代码 

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        int m = s1.size(), n = s2.size();

        vector<vector<int>> dp(m + 1, vector<int>(n + 1));

        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++) {
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                if (s1[i - 1] == s2[j - 1])
                    dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + s1[i - 1]);
            }

        int sum = 0;
        for (auto s : s1)
            sum += s;
        for (auto s : s2)
            sum += s;

        return sum - 2 * dp[m][n];
    }
};

原文地址：https://blog.csdn.net/2401_82677021/article/details/144407632

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