题目链接：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

这道题的要求是计算两个单词的距离，即从word1转换到word2所需要的最少步数。可以有三种操作：插入字符、删除字符、替换字符。

计算单词间的距离，令d[i, j]表示word1的前i个字符与word2的前j个字符之间的距离。则d[i, j]的值有如下几种情况：

  1. 如果word1[i-1]等于word2[j-1]，则表明该位置不需要操作，即d[i, j] = d[i-1, j-1]；
  2. 如果word1[i-1]不等于word2[j-1]，则可能需要删除或添加或替换字符。如果需要删除或添加字符（删除和添加实为相反操作，这里只考虑删除吧）。可以选择删除word1[i-1]，这样d[i, j]就是word1的前i-1位和word2的前j位的距离再加1步删除操作，即d[i, j] = d[i-1, j] + 1。如果删除word1[j-1]，则d[i, j] = d[i, j-1] + 1。当然，如果替换，则d[i, j] = d[i-1, j-1] + 1。因此，d[i, j] = min(d[i-1, j], d[i, j-1], d[i-1, j-1]) + 1。

时间复杂度：O(mn)

空间复杂度：O(mn)

 1 class Solution
 2 {
 3 public:
 4     int minDistance(string word1, string word2)
 5     {
 6         int l1 = word1.size(), l2 = word2.size();
 7         
 8         if(l1 == 0 || l2 == 0)
 9             return max(l1, l2);
10         
11         vector<vector<int> > vvi(l1 + 1, vector<int>(l2 + 1, 0));
12         
13         for(int i = 0; i <= l1; ++ i)
14             vvi[i][0] = i;
15         for(int j = 0; j <= l2; ++ j)
16             vvi[0][j] = j;
17         
18         for(int i = 1; i <= l1; ++ i)
19             for(int j = 1; j <= l2; ++ j)
20                 if(word1[i - 1] == word2[j - 1])
21                     vvi[i][j] = vvi[i - 1][j - 1];
22                 else
23                     vvi[i][j] = min(min(vvi[i - 1][j], vvi[i][j - 1]), vvi[i - 1][j - 1]) + 1;
24         
25         return vvi[l1][l2];
26     }
27 };

这里只是简单实现了动态规划的思路，如想详细研究，请看这pdf。