Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

  1. Insert a character
  2. Delete a character
  3. Replace a character

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation: 
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation: 
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

题解：

想到很像之前752的一道广度优先搜索，但是搜索范围太大，应该是动态规划问题。

如果两个字符相等，那dp[i][j] = dp[i - 1][j - 1]，不相等的话就等于dp[i][j] = min(dp[i - 1][j - 1], min(dp[i][j - 1], dp[i - 1][j])) + 1，分别是更改/插入/删除的情况。

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n = word1.length(), m = word2.length();
        vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= m; j++) {
                dp[i][j] = max(i, j);
            }
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; 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 - 1], min(dp[i][j - 1], dp[i - 1][j])) + 1;
                }
            }
        }
        return dp[n][m];
    }
};