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
用m*n的矩阵vector中的（i，j）存储从word1中(0...i)到word2中（0...j）的最小变化次数
矩阵中，从上到下对应删除操作，从左到右对应插入操作，从左上到右下对应Replace操作。
class Solution { public: int minDistance(string word1, string word2) { //1->2 if(word1==word2) return 0; int len1 = word1.size(); int len2 = word2.size(); vector<int> temp(len2+1,0); vector<vector<int> > vec(len1+1,temp); for(int i=1;i<=len1;i++){ vec[i][0] = i; } for(int j=1;j<=len2;j++){ vec[0][j] = j; } for(int i=1;i<=len1;i++){ for(int j=1;j<=len2;j++){ int tem1,tem2; tem1 = min(vec[i-1][j]+1,vec[i][j-1]+1); tem2 = word1[i-1]==word2[j-1] ? vec[i-1][j-1]:vec[i-1][j-1]+1; vec[i][j] = min(tem1,tem2); }//end for }//end for return vec[len1][len2]; } };
[LeetCode] Edit Distance（很好的DP）,布布扣,bubuko.com
[LeetCode] Edit Distance（很好的DP）
原文：http://www.cnblogs.com/Xylophone/p/3919112.html