leetcode 583 两个字符串的删除操作

给定两个单词 word1 和 word2，找到使得 word1 和 word2 相同所需的最小步数，每步可以删除任意一个字符串中的一个字符。

示例：

输入: “sea”, “eat”
输出: 2
解释: 第一步将"sea"变为"ea"，第二步将"eat"变为"ea"

解法1：
动态规划
链接：https://leetcode-cn.com/problems/delete-operation-for-two-strings/solution/liang-chong-jie-fa-by-jason-2-34/

    int minDistance(string word1, string word2) {
        const int len1 = word1.size();
        const int len2 = word2.size();
        vector<vector<int>> d(len1+1,vector<int>(len2+1,0));
        for(int i=0;i<=len1;++i){
            for(int j=0;j<=len2;++j){
                if(i == 0 && j == 0){
                    d[i][j] = 0;
                }else if(i == 0 && j){
                    d[i][j] = j;
                }else if(i && j == 0){
                    d[i][j] = i;
                }else{
                    d[i][j] = min(
                        min(d[i-1][j],d[i][j-1])+1,
                        d[i-1][j-1] + (word1[i-1] == word2[j-1] ? 0 :2)
                    );
                }
            }
        }
        return d[len1][len2];
    }

解法2：
最长公共字串
链接：https://leetcode-cn.com/problems/delete-operation-for-two-strings/solution/liang-chong-jie-fa-by-jason-2-34/

    int minDistance(string word1, string word2) {
        const int len1 = word1.size();
        const int len2 = word2.size();
        vector<vector<int>> d(len1+1,vector<int>(len2+1,0));
        for(int i=0;i<=len1;++i){
            for(int j=0;j<=len2;++j){
                if(i == 0 || j == 0){
                    d[i][j] = 0;
                }else{
                    if(word1[i-1] == word2[j-1]) {
                        d[i][j] = d[i-1][j-1] + 1;
                    }else d[i][j] = max(d[i-1][j],d[i][j-1]);
                }
            }
        }
        return len1+len2-2*d[len1][len2];
    }