LeetCode动态规划专题：第十九天

Shirandexiaowo

于 2021-09-07 15:27:03 发布

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分类专栏：动态规划文章标签： leetcode 动态规划

本文链接：https://blog.youkuaiyun.com/MATLAB2020ab/article/details/120155421

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本文介绍了LeetCode中的三个经典问题：判断字符串是否为子序列，求两个字符串的最长公共子序列，以及计算两个字符串的最小编辑距离。通过实例代码展示了如何用C++实现这些算法，适用于基础的字符串操作和动态规划技巧。

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Leetcode392. 判断子序列

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int k = 0;
        for (auto c : t)
            if (k < s.size() && c == s[k])
                k ++ ;
        return k == s.size();
    }
};

Leetcode1143. 最长公共子序列

class Solution {
public:
    int longestCommonSubsequence(string s1, string s2) {
        int n = s1.size(), m = s2.size();
        s1 = ' ' + s1;
        s2 = ' ' + s2;
        vector<vector<int>> f(n + 1, vector<int>(m + 1));
        for (int i = 1; i <= n; i ++ )
            for (int j = 1; j <= m; j ++ )
            {
                f[i][j] = max(f[i - 1][j], f[i][j - 1]);
                if (s1[i] == s2[j]) 
                    f[i][j] = max(f[i][j], f[i - 1][j - 1] + 1);
            }
        return f[n][m];
    }
};

Leetcode72. 编辑距离

class Solution {
public:
    int minDistance(string s1, string s2) {
        int n = s1.size(), m = s2.size();
        s1 = ' ' + s1;
        s2 = ' ' + s2;
        vector<vector<int>> f(n + 1, vector<int>(m + 1));
        for (int j = 0; j <= m; j ++ ) f[0][j] = j;
        for (int i = 1; i <= n; i ++ ) f[i][0] = i;

        for (int i = 1; i <= n; i ++ )
            for (int j = 1; j <= m; j ++ ) {
                f[i][j] = min(f[i - 1][j], f[i][j - 1]) + 1;
                int t = s1[i] != s2[j];
                f[i][j] = min(f[i][j], f[i - 1][j - 1] + t);
            }
        return f[n][m];
    }
};