Edit Distance

leetcode的题目

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

 

这是一个动态规划的题目

wiki上的递推公式如下

代码如下

 1 　　int minDistance(string word1, string word2) {
 2         int **dp = new int*[word1.length() + 1];
 3         for(int i = 0; i < word1.length() + 1; i ++)
 4             dp[i] = new int[word2.length() + 1];
 5 
 6         for(int i = 0; i < word1.length() + 1; i ++)
 7             dp[i][0] = i;
 8         for(int i = 0; i < word2.length() + 1; i ++)
 9             dp[0][i] = i;
10         
11         for(int i = 1; i < word1.length() + 1; i ++)
12             for(int j = 1; j < word2.length() + 1; j ++)
13             {
14                 dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1;//insert or delete
15 
16                 if(word1[i - 1] != word2[j - 1])//modify
17                 {
18                     dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + 1);
19                 }
20                 else
21                 {
22                     dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
23                 }
24             }
25 
26             return dp[word1.length()][word2.length()];
27     }

 时间复杂度和空间复杂度都是O(M*N)

如果不要求计算修改过程, 那么空间复杂度可以优化到线性

 

感觉自己对动态规划还是很不敏感

 

查阅相关资料后发现, 这个题目属于文本比较算法, 还有很多相关的内容, 比如说LCS和一些其他更加高级的算法, 以后有空再研究一下

posted @ 2014-08-21 21:09  qeDVuHG  阅读(164)  评论(0)  编辑  收藏  举报