[leetcode] 72 Edit Distance
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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
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题目大致可以理解为字符串word1经过几步变化可以变为word2,添加、删除、调换均算一次操作
如:“abs”--->"kas" 就需要两步 abs-->bas-->kas
leetcode给的提示是动态规划,所以我们也就顺着这个思路写dp算法(其实之前自己想过分析两个string中相同字符的位置关系,不过似乎太复杂了点)
假设word1的长度为m,word2的长度为n,建立二位数据dp[m+ 1][n+ 1],dp[i][j]表示word1的前i位于word2的前j位的最短距离
dp[i][j]= min( min(dp[i- 1][j], dp[i][j- 1])+ 1 , dp[i- 1][j- 1]+ flag ) 其中如果word1[i]和word2[j]相同则flag= 0,否则= 1;
class Solution {public: int minDistance(string word1, string word2) { int m= word1.length(); int n= word2.length(); if(m== 0) return n; if(n== 0) return m; vector<vector<int> > dp(m+ 1, vector<int>(n+ 1, 0)); for(int i= 1; i< m+ 1; i++){ dp[i][0]= i; } for(int i= 1; i< n+ 1; i++){ dp[0][i]= i; } for(int i= 1; i< m+ 1; i++){ for(int j= 1; j< n+ 1; j++){ int flow= 0; if(word1[i- 1]!= word2[j- 1]) flow= 1; dp[i][j]= min(min(dp[i- 1][j], dp[i][j- 1])+ 1, dp[i- 1][j- 1]+ flow); } } return dp[m][n]; }};
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