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文章目录
- 一、题目
- 二、题解
一、题目
Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.
You have the following three operations permitted on a word:
Insert a character
Delete a character
Replace a character
Example 1:
Input: word1 = “horse”, word2 = “ros”
Output: 3
Explanation:
horse -> rorse (replace ‘h’ with ‘r’)
rorse -> rose (remove ‘r’)
rose -> ros (remove ‘e’)
Example 2:
Input: word1 = “intention”, word2 = “execution”
Output: 5
Explanation:
intention -> inention (remove ‘t’)
inention -> enention (replace ‘i’ with ‘e’)
enention -> exention (replace ‘n’ with ‘x’)
exention -> exection (replace ‘n’ with ‘c’)
exection -> execution (insert ‘u’)
Constraints:
0 <= word1.length, word2.length <= 500
word1 and word2 consist of lowercase English letters.
二、题解
class Solution {
public:int minDistance(string word1, string word2) {int n = word1.size(),m = word2.size();vector<vector<int>> dp(n+1,vector<int>(m+1,0));for(int i = 1;i <= n;i++) dp[i][0] = i;for(int j = 1;j <= m;j++) dp[0][j] = j;for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++){if(word1[i-1] == word2[j-1]) dp[i][j] = dp[i-1][j-1];else dp[i][j] = min(dp[i-1][j] + 1,min(dp[i][j-1] + 1,dp[i-1][j-1] + 1));}}return dp[n][m];}
};
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