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【LeetCode】Edit Distance

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];        }    }};

【LeetCode】Edit Distance