编辑距离

给出两个单词word1和word2，计算出将word1 转换为word2的最少操作次数。

你总共三种操作方法：

• 插入一个字符
• 删除一个字符
• 替换一个字符
  

您在真实的面试中是否遇到过这个题？

Yes

样例

给出 work1="mart" 和 work2="karma"

class Solution {public:    /*     * @param word1: A string     * @param word2: A string     * @return: The minimum number of steps.     */    int minDistance(string &word1, string &word2) {        // write your code here        int m = word1.size();        int n = word2.size();        if (m == 0) {            return n;        } else if (n == 0) {            return m;        }        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));        dp[0][0] = 0;        for (int i = 1; i <= m; i++) {            dp[i][0] = i;        }        for (int j = 1; 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];                } else {                    dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1;                    dp[i][j] = min(dp[i - 1][j - 1] + 1, dp[i][j]);                }            }        }        return dp[m][n];    }};