标签：style blog http io ar color sp for strong
Edit Distance
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
递归超时。
又是一个“走地图”的动态规划算法。
word1, word2只有两个字符串，因此可以展平为一个二维地图，转换代价即从左上角走到右下角的最小代价。
记代价矩阵为dist. dis[i][j]代表word1[0,i)转换为word2[0,j)的最小代价。
当word1到达第i-1个元素，word2到达第j-1个元素:
地图上往右一步代表word1当前位置插入了word2[j-1]字符，dis[i][j]=dis[i][j-1]+1, 下一次比较word1[i-1]和word2[j]
地图上往下一步代表word1当前位置删除了word1[i-1]字符，dis[i][j]=dis[i-1][j]+1, 下一次比较word1[i]和word2[j-1]
地图上往对角线一步代表word1[i-1]替换为word2[j-1]（相同则不替换），dis[i][j]=dis[i-1][j-1]+((word1[i-1]==word2[j-1])?0:1),下一次比较word1[i]和word2[j]
示例：word1="a", word2="ab"
dis 0 a b
0 0 1 2
a 1 0 1
class Solution { public: int minDistance(string word1, string word2) { if(word1 == "" && word2 == "") return 0; else if(word1 == "" && word2 != "") return word2.size(); else if(word1 != "" && word2 == "") return word1.size(); else { int size1 = word1.size(); int size2 = word2.size(); vector<vector<int> > dis(size1+1, vector<int>(size2+1, 0)); for(int i = 0; i <= size1; i ++) dis[i][0] = i; for(int j = 0; j <= size2; j ++) dis[0][j] = j; for(int i = 1; i <= size1; i ++) { for(int j = 1; j <= size2; j ++) { //move right from dis[i][j-1], insert word2[j-1] int dis1 = dis[i][j-1] + 1; //move down from dis[i-1][j], delete word1[i-1] int dis2 = dis[i-1][j] + 1; //move diagonal, replace word1[i-1] with word2[j-1] int replace = (word1[i-1]==word2[j-1])?0:1; int dis3 = dis[i-1][j-1] + replace; dis[i][j] = min(dis1, min(dis2,dis3)); } } return dis[size1][size2]; } } };
标签：style blog http io ar color sp for strong
原文地址：http://www.cnblogs.com/ganganloveu/p/4153464.html
