Edit Distance

Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

分析：

处理这道题也是用动态规划。

动态数组dp[word1.length+1][word2.length+1]

dp[i][j]表示从word1前i个字符转换到word2前j个字符最少的步骤数。

假设word1现在遍历到字符x，word2遍历到字符y（word1当前遍历到的长度为i，word2为j）。

以下两种可能性：

1. x==y，那么不用做任何编辑操作，所以dp[i][j] = dp[i-1][j-1]

2. x != y

(1) 在word1插入y， 那么dp[i][j] = dp[i][j-1] + 1

(2) 在word1删除x， 那么dp[i][j] = dp[i-1][j] + 1

(3) 把word1中的x用y来替换，那么dp[i][j] = dp[i-1][j-1] + 1

最少的步骤就是取这三个中的最小值。

最后返回 dp[word1.length+1][word2.length+1] 即可。

class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size();
        int n = word2.size();
        vector<vector<int>> dp(m+);
        for(size_t i = ; i<m+; i++)
            for(size_t j = ; j<n+; j++)
                dp[i].push_back(-);
        for(size_t i=; i<m+;i++)
            dp[i][] =i;
        for(size_t j=; j<n+; j++)
            dp[][j] =j;
        for(int i=; i<m+; i++)
            for(int j =; j<n+; j++)
            {
                if(word1[i-] == word2[j-]){
                    dp[i][j] = dp[i-][j-];
                }
                else{
                    int insert = dp[i-][j]+;
                    int replace = dp[i-][j-]+;
                    int del = dp[i][j-]+;
                    dp[i][j] = min(del,min(insert, replace));
                }
            }
        return dp[m][n];
    }
};