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leetcode72. Edit Distance(编辑距离)

以下为个人翻译方便理解编辑距离问题是一个经典的动态规划问题.首先定义dp[i][j表示word1[0..i-1]到word2[0..j-1]的最小操作数(即编辑距离). 状态转换方程有两种情况:边界情况和一般情况,以上表示中 i和j均从1开始(注释:即至少一个字符的字符串向一个字符的字符串转换,0字符到0字符转换编辑距离自然为0) 1.边界情况:将一个字符串转化为空串,很容易看出把word[0...i-1]转化成空串""至少需要i次操作(注释:i次删除),则其编辑距离为i,即:dp[…

[leetcode72]Edit Distance（dp）

题目链接:https://leetcode.com/problems/edit-distance/ 题意:求字符串的最短编辑距离,就是有三个操作,插入一个字符.删除一个字符.修改一个字符,最终让两个字符串相等. DP,定义两个字符串a和b,dp(i,j)为截至ai-1和bj-1时的最短编辑距离. 当ai-1=bi-1的时候,有dp(i,j)=min(dp(i,j),dp(i-1,j-1)),对应不做任何操作: 不相等的时候会有dp(i,j)=min(dp(i,j),dp(i-1,j-1)+1),…

leetcode72. 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) Repla…

[Swift]LeetCode72. 编辑距离 | Edit Distance

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2. You have the following 3 operations permitted on a word: Insert a character Delete a character Replace a character Example 1: Input: word1 = "h…

class Solution { private: ][]; public: int minDistance(string word1, string word2) { int len1 = word1.length(); int len2 = word2.length(); ;i<=len1;i++) d[i][]= i; ;j<=len2;j++) d[][j]=j; ;i <=len1;i++) { ;j<=len2;j++) { int diff; ] == word2[j…

Java解决LeetCode72题 Edit Distance

题目描述地址 : https://leetcode.com/problems/edit-distance/description/ 思路使用dp[i][j]用来表示word1的0~i-1.word2的0~j-1的最小编辑距离我们可以知道边界情况:dp[i][0] = i.dp[0][j] = j,代表从 "" 变为 dp[0~i-1] 或 dp[0][0~j-1] 所需要的次数同时对于两个字符串的子串,都能分为最后一个字符相等或者不等的情况: 如果word1[i-1] == w…

动态规划两个字符串之间的编辑距离 leetcode72

public static int minDistance(String word1, String word2) { char[] s1 = word1.toCharArray(); char[] s2 = word2.toCharArray(); int len1 = s1.length; int len2 = s2.length; int N = Math.max(len1, len2); int[][] d = new int[N + 1][N + 1]; for (int i = 0;…