1classSolution {2publicintminDistance(String word1, String word2) {3intlen1 =word1.length();4intlen2 =word2.length();56if(len1 == 0)returnlen2;7if(len2 == 0)returnlen1;89int[] dp =newint[len1+1];1011for(intj = 0; j <= len1; j++) {12dp[j] =j;13}1415for(inti =...
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...
if (word1[M - 1] == word2[N - 1]) { return minDistance(word1.substr(0, M - 1), word2.substr(0, N - 1)); } return 1 + min(min(minDistance(word1.substr(0, M - 1), word2), minDistance(word1, word2.substr(0, N - 1))), minDistance(word1.substr(0, M - 1),...
int deleteMin = subMinDistance(i1 + 1, i2, s1, s2, memo) + 1; int changeMin = subMinDistance(i1 + 1, i2 + 1, s1, s2, memo) + 1; int insertMin = subMinDistance(i1, i2 + 1, s1, s2, memo) + 1; memo[i1][i2] = Math.min(deleteMin, Math.min(changeMin, insertMin)...
[刷算法DAY7] 動態規劃練習 | LeetCode 96. Unique Binary Search Trees ___Null 54 0 算法,成为程序员强者的必经之路! ACM金牌大牛授课 [刷算法DAY2] 動態規劃練習 | LeetCode 64. Minimum Path Sum ___Null 29 0 [刷算法DAY4] 動態規劃練習 | LeetCode 343. Integer Break ___Null 46 0 ...
删除一个字符 替换一个字符 示例1: 输入:word1 = "horse", word2 = "ros"输出:3解释:horse -> rorse (将 'h' 替换为 'r') rorse -> rose (删除 'r') rose -> ros (删除 'e') 示例2: 输入:word1 = "intention", word2 = "execution"输出:5解释:intention -> inention (删除 't')...
1、题目描述 Given two strings S and T, determine if they are both one edit distance apart. 2、问题描述: 这道题是之前那道Edit Distance的拓展,然而这道题并没有那道题难,这道题只让我们判断两个字符串的编辑距离是否为1,那么我们只需分下列三种情况来考虑就行了: ...
publicintminDistance(Stringword1,Stringword2){if((word1==null||word2==null)&&(word1.length()==0||word2.length()==0))return0;if(word1==null||word1.length()==0)returnword2.length();if(word2==null||word2.length()==0)returnword1.length();intresult=0;int[][]dp=newint[word...
72. Edit Distancehttps://leetcode.com/problems/edit-distance/class Solution:def minDistance(self, word1: str, word2: str) -> int: rows=len(word1)+1 col=len(word2)+1 dp=[] for i in range(rows): m=[] for j in range(col): ...
Leetcode[161] One Edit Distance LeetCode[161] One Edit Distance Given two strings S and T, determine if they are both one edit distance apart. String 复杂度 O(N),O(1) 思路 考虑如果两个字符串的长度 > 1,是肯定return false; 当两个字符串中有不同的字符出现的时候,说明之后的字符串一定要...