leetcode习题解答：72. Edit Distance

难度：Hard

描述：

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

思路：课上讲过，复习一下。

需要找出word1到word2的编辑距离。

dp[i][j]状态转移如下情况：

1.从dp[i-1][j-1]转移而来，如果最新的两个字符相同，就不需要替换字符，dp[i-1][j-1]+0，否则+1

2.从dp[i-1][j]转移而来，为了处理加上的这个字符，dp[i-1][j]+1   对应的操作是word1串长度不够，要+1

3.从dp[i][j-1]转移而来，为了处理加上的这个字符，dp[i][j-1]+1   对应的操作是word1串长度过长，要-1

dp[i][j] = min{1.，2.，3.}

如果画一个表格来填写，模拟这个过程，可以清楚看到算法的意图。

代码：

class Solution {public:    int minDistance(string word1, string word2) {        int dp[1000][1000];        int m = word1.size();        int n = word2.size();        for (int i = 0; i <= m; i++){            dp[i][0] = i;        }        for (int i = 0; i <= n; i++){            dp[0][i] = i;        }        for (int i = 1; i <= m; i++){            for (int j = 1; j<=n; j++){                int min = 99999;                int bonus = 0;                if (word1[i-1] != word2[j-1]) bonus = 1;                if (dp[i-1][j-1] < min) min = dp[i-1][j-1]+bonus;                if (dp[i][j-1] < min) min = dp[i][j-1]+1;                if (dp[i-1][j] < min) min = dp[i-1][j]+1;                dp[i][j] = min;            }        }        return dp[m][n];    }};