动态规划——72. Edit Distance［HARD］

题目描述

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

这是编辑距离，两个字符串，利用最少的 增、删、改 变成同一个字符串。

解题思路

1）dp［i］［j］：word1 0～i 和 word2 0～j 的编辑距离。

2）考虑word1［i］ word2［j］的对齐方式：

注意因为出现 i－1，j－1，所以要考虑数组越界问题，因此多生成0行和0列，对应某一个str为空时的情况

代码如下

class Solution {public:  int minDistance(string word1, string word2) {if (word1.empty() && word2.empty())return 0;if (word1.empty())return word2.size();if (word2.empty())return word1.size();vector<int> tmp(word2.size()+1, 0);vector<vector<int> > dp(word1.size()+1, tmp);for (int i = 0; i <= word2.size(); i++)dp[0][i] = i;for (int j = 0; j <= word1.size(); j++)dp[j][0] = j;for (int i = 1; i <= word1.size(); i++) {for (int j = 1; j <= word2.size(); j++) {int op = (word1[i-1] == word2[j-1] )? dp[i - 1][j - 1] : dp[i - 1][j - 1] + 1;dp[i][j] = fmin(dp[i - 1][j] + 1, fmin(dp[i][j - 1] + 1, op));}}return dp[word1.size()][word2.size()];}};