leetcode:72. 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

题意&解题思路：

给定两个字符串，求最短编辑距离，三种操作，增加一个字符，删除一个字符，替换一个字符。

用DP解决，设 dp[i][j] 表示将 word1[0 ~ i-1] 转换为 word2[0 ~ j-1] 所需的最小操作数。对 word1[0 ~ i-1] 而言，转换为 "" (空字符串)的最小编辑距离为i，同理，对 word2[0 ~ j-1] 而言，转换为 "" (空字符串)的最小编辑距离为j。故有：

dp[i][0] = i;

dp[0][j] = j;

对一般的 word1[0 ~ i-1] 和 word2[0 ~ j-1]，若前面 word1[0 ~ i-2] 和 word2[0 ~ j-2] 的最小编辑距离已知，则有：

1.当 word1[0 ~ i-1] == word2[0 ~ j-1] 时，有dp[i][j] = dp[i-1][j-1]；

2.当 word1[0 ~ i-1] != word2[0 ~ j-1]时，可以有三种选择

1)将 word1[i-1] 替换为字符 word2[j-1]，则 dp[i][j] = dp[i-1][j-1] + 1

2)将字符word1[i-1]删除后有 word1[i-2] == word2[j-1]，则 dp[i][j] = dp[i-1][j] + 1

3)将字符word2[j-1]增加到 word1[i-1] 位置使得word1[i-1] == word2[j-1]，则 dp[i][j] = dp[i][j-1] + 1

选择最短编辑距离，亦即 dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1])+1。

根据上述思路，代码如下：

class Solution {public:    int minDistance(string word1, string word2) {        int l1 = word1.length(), l2 = word2.length();        if(l1 == 0)return l2;        if(l2 == 0)return l1;        int dp[l1+1][l2+1];        memset(dp, 0,sizeof(dp));        for(int i = 0; i <= l1; i++)dp[i][0] = i;        for(int i = 0; i <= l2; i++)dp[0][i] = i;        for(int i = 1; i <= l1; i++)            for(int j = 1; j <= l2; 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] + 1, min(dp[i-1][j] + 1, dp[i][j-1] + 1));            }        return dp[l1][l2];    }};