[Leetcode] 161. One Edit Distance 解题报告 题目: Given two strings S and T, determine if they are both one edit distance apart. 思路: One Edit Distance意味着一个字符串s可以通过删除,增加或者修改其中的一个字符变成另外一个字符串t。因此我们可以根据s和t的长度分别处理:如果两者长度相同,则判断...
Difficulty: Medium More:【目录】LeetCode Java实现 回到顶部 Description Given two strings S and T, determine if they are both one edit distance apart. 回到顶部 Intuition 同时遍历比较S和T的字符,直至遇到不同的字符:如果S和T的字符个数相同时,跳过不同的那个字符,继续遍历;如果S和T的字符个数相差为1...
{if(s[i] ==t[i])continue;returns.substr(i) == t.substr(i+1); }returntrue;//说明前面的都相等只有最后一个不匹配}boolisSameLen(strings,stringt) {intcnt =0;for(inti =0; i < s.size(); i++) {if(s[i] !=t[i]) cnt++; }returncnt ==1; }boolisOneEditDistance(strings,strin...
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...
LeetCode 161. One Edit Distance (Java版; Medium) 题目描述 AI检测代码解析 Given two strings s and t, determine if they are both one edit distance apart. Note: There are 3 possiblities to satisify one edit distance apart: Insert a character into s to get t ...
Leetcode: One Edit Distance 注意这道题:Must be exactly one distance apart. Not the same. 1publicclassSolution {2publicbooleanisOneEditDistance(String s, String t) {3for(inti=0; i<Math.min(s.length(), t.length()); i++) {4if(s.charAt(i) !=t.charAt(i)) {5if(s.length() ==...
Provide all my solutions and explanations in Chinese for all the Leetcode coding problems. Same as this: LeetCode All in One 题目讲解汇总(持续更新中...) Note: All explanations are written in Github Issues, please do not create any new issue or pull request in this project since the proble...
One Edit Distance.md update Oct 17, 2019 Leetcode-162. Find Peak Element.md update Oct 9, 2019 Leetcode-164. Maximum Gap.md update Sep 19, 2019 Leetcode-165. Compare Version Numbers.md update Sep 15, 2019 Leetcode-168. Excel Sheet Column Title.md update Sep 17, 2019 Leetcode-169....
Given two strings S and T, determine if they are both one edit distance apart. 思路: 两个字符串是否只通过一步变换(替换、删除、增加)就变成相同字符串。 如果字符串长度差大于1,肯定是不可能。 如果两个字符串本来就相同,也不可能。 可能的情况只存在于字符串长度相同但字符串不相同或长度差为1。
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: ...