标签：style blog http color io for ar 2014
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
标签：style blog http color io for ar 2014
原文地址：http://www.cnblogs.com/stevenczp/p/3928087.html
