算法训练营第四十三天（9.5）| 动态规划Part16：距离

目录

 Leecode 583.两个字符串的删除操作

 Leecode 72.编辑距离

 Leecode 583.两个字符串的删除操作

题目地址：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

题目类型：最长子序列

class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size(), n = word2.size();
        // dp[i][j]代表word1从0....i-1和word2从0......j-1之间最少的步数
        vectorint>> dp(m + 1, vector<int>(n + 1));
        // 初始化
        for (int i = 0; i <= m; ++i) dp[i][0] = i;
        for (int j = 0; j <= n; ++j) dp[0][j] = j;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (word1[i - 1] == word2[j - 1]) dp[i][j] = dp[i - 1][j - 1];
                // 因为要模拟删除元素的操作，所以要加一
                else dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
            }
        }
        return dp[m][n];
    }
};

 Leecode 72.编辑距离

题目地址：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

题目类型：最长子序列

class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size(), n = word2.size();
        vectorint>> dp(m + 1, vector<int>(n + 1));
        for (int i = 0; i <= m; ++i) dp[i][0] = i;
        for (int j = 0; j <= n; ++j) dp[0][j] = j;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                // 如果相同则不需要修改
                if (word1[i - 1] == word2[j - 1]) dp[i][j] = dp[i - 1][j - 1];
                else dp[i][j] = min({dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]}) + 1;
            }
        }
        return dp[m][n];
    }
};