Edit Distance

一. 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

Difficulty:Hard

TIME：42MIN

解法（动态规划）

刚开始看到这道题，确实一时间没有想到该怎么解，这道题其实和求两个字符串的最长公共子序列类似，动态规划的思想也十分相似，知道了最优子结构，其实这道题也就没什么难度了。

/*简单来说，就是把插入，删除和替换对应于某种状态，如果dp表示长度为i的子串到长度为j的子串的编辑距离，那么最优子结构表示如下：     如果dp[i - 1] == dp[j - 1]，那么dp[i][j] = dp[i-1][j-1];     如果dp[i - 1] != dp[j - 1]，那么对应于三种情况:      1) dp[i][j] = dp[i - 1][j - 1] + 1 替换     2) dp[i][j] = dp[i - 1][j] + 1 删除     3) dp[i][j] = dp[i][j - 1] + 1 插入*/int minDistance(string word1, string word2) {    vector<vector<int>> dp(word1.size() + 1,vector<int>(word2.size() + 1, 0));    for(int i = 0; i <= word2.size(); i++)        dp[0][i] = i;    for(int i = 0; i <= word1.size(); i++)        dp[i][0] = i;    for(int i = 1; i <= word1.size(); i++) {        for(int j = 1; j <= word2.size(); 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 - 1][j - 1] + 1, dp[i][j - 1] + 1));        }    }    return dp[word1.size()][word2.size()];}

代码的时间复杂度为O(nm)。