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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
思路：
这个去年学算法的时候学过，当时觉得好难啊，现在一看，这题真简单。就是一道常规的动态规划题目。直接AC，高兴~~
用dp[m][n]存储word1的前m个字符与word2的前n个字符的最小匹配距离。
那么dp[i][j] = dp[i-1][j]+1 （word1删除一个字符）、dp[i][j-1] + 1 （word2删除一个字符）、 dp[i-1][j-1] + ((word1[i-1]==word2[j-1]) ? 0（当前字符相等） : 1（替换）)) 中最小的值。
class Solution { public: int minDistance(string word1, string word2) { int len1 = word1.length(); int len2 = word2.length(); vector<vector<int>> dp(len1 + 1, vector<int>(len2 + 1, 0)); for(int i = 1; i < len1 + 1; i++) { dp[i][0] = i; } for(int j = 1; j < len2 + 1; j++) { dp[0][j] = j; } for(int i = 1; i < len1 + 1; i++) { for(int j = 1;j < len2 + 1; j++) { dp[i][j] = min(min(dp[i-1][j] + 1, dp[i][j-1] + 1), dp[i-1][j-1] + ((word1[i-1]==word2[j-1]) ? 0 : 1)); } } return dp[len1][len2]; } };
【leetcode】Edit Distance (hard)
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原文地址：http://www.cnblogs.com/dplearning/p/4180430.html
