leetcode 72. Edit Distance

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:

Insert a character
Delete a character
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')

题意：一个字符串，插入，删除，或替换变道另一个字符串，输出最小操作数。

思路：dp.
可以抽象成一个矩阵。
比如下面矩阵表示，比如第1行第3列，表示由空变到ab需要的操作数。第3行第1列表示由ac变到空需要的操作数。

  空 a b c
空 0 1 2 3
a  1 0 ?
c  2
c  3

dp[i][j]表示，字符串s的前i个字母变到字符串t的前j个字母需要的最少的操作数。那么dp[i][j] = min(dp[i-1][j-1],dp[i-1][j],dp[i][j-1]) + 1.
分别对应替换，删除，添加。
比如问好位置，可由空->aa替换后一个a,或者a->a添加一个b,或则空->aab删除个a.

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