[leetcode]Edit Distance
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Edit Distance
Difficulty:Hard
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
题目就是求编辑距离,这个课上也讲过了,就是用动态规划的算法来解决,通过构造编辑距离矩阵来计算最短的编辑距离。
大概就是这样的:
矩阵中每一个数字表示在该位置相应的两个字符串的前i和前j个字符的编辑距离,一开始只有第一行和第一列有值,就是上图的黑色数字,红色的数字就是要算出来的,可以根据(i-1,j)(i,j-1)(i-1,j-1)三个位置的值算出来。
算的方法就是E(i,j)min{E(i-1,j)+1,E(i,j-1)+1,E(i-1,j-1)+diff(s1[i],s2[j])};s1,s2就是要算编辑距离的两个字符串,diff表示两个字符的编辑距离,相等则为0,不等为1 。
算完后,最右下角的值就是两个字符串的编辑距离。
看了很多编辑距离的算法都跟这个差不多,不过实现起来都很不一样,我这个还是比较慢的。。。
int minDistance(string word1, string word2) { int row = word1.size(); int col = word2.size();vector<vector<int>> allDistance(row + 1, *(new vector<int>(col + 1, 0)));//编辑距离的矩阵for (int i = 0; i < allDistance[0].size(); i++){//初始化第一行allDistance[0][i] = i;}for (int i = 0; i < allDistance.size(); i++){//初始化第一列allDistance[i][0] = i;}for (int i = 1; i < allDistance.size(); i++){for (int j = 1; j < allDistance[0].size(); j++){int disLeft = allDistance[i][j - 1] + 1;//E(i,j-1)+1int disUp = allDistance[i - 1][j] + 1;//E(i-1,j)+1int disDiag = allDistance[i - 1][j - 1] + (word1.at(i - 1) == word2.at(j - 1) ? 0 : 1);//E(i-1,j-1)+diff(s1[i],s2[j])int disMin = min(disLeft, disUp);disMin = min(disMin, disDiag);allDistance[i][j] = disMin;}}return allDistance[allDistance.size() - 1][allDistance[0].size() - 1];}
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