LeetCode:95.编辑距离

目录

1.编辑距离

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1.编辑距离

dp[i][j]表示word1的前i个字符变为word2的前j个字符所需要的最小操作数

对于插入来说相当于在i+1位置插入一个字符，使i+1的字符等于j的字符，所以只需要让前i个字符转化为前j-1个字符的操作数加1即可，dp[i][j] = dp[i][j - 1] + 1

对于删除来说，相当于删除第i个字符，使前i-1个字符变为前j个字符，dp[i][j] = dp[i - 1][j] + 1

对于替换来说，将第i个字符替换使其与第j个字符相等，dp[i][j] = dp[i - 1][j - 1] + 1

class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size(), n = word2.size();
        if(m * n == 0) return m + n;
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));
        word1 = " " + word1, word2 = " " + word2;
        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] == word2[j]) dp[i][j] = dp[i - 1][j - 1];
                else
                {
                    int x = dp[i][j - 1] + 1, y = dp[i - 1][j] + 1, z = dp[i - 1][j - 1] + 1;
                    dp[i][j] = min(x, min(y, z));
                }
            }
        return dp[m][n];
    }
};