}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,stringt) {if(abs(s.size() - t.size()) >1)returnfalse;if(s.size() ...
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时,跳过较长的字符串的当天字符,继续遍历。如果剩下的字符都相等,那么返...
LeetCode 161 One Edit Distance Problem: Given two strings S and T, determine if they are both one edit distance apart. Analyse: delete, add, replace 都是one edit distance. delete 和 add: 两个string 的长度相差一 replace: 两个string的长度相同 伪代码: str......
abc t的长度比较大的时候,说明t.substring(i + 1).equals(s.substring(i)), 才能保证距离为1。 abe ace 当剩下的字符串距离相等的时候,说明只有当剩下的字符串相等的时候,才能保证距离为1. 代码 public boolean isOneEditDistance(String s, String t) { for(int i = 0;i < Math.min(s.length(),...
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 Delete a character from s to get t Replace a character of s to get t ...
【LeetCode】 -- One Edit Distance 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...
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() ==...
class Solution: def isOneEditDistance(self, s: str, t: str) -> bool: n,m=len(s),len(t) flag=False if n==m: for i in range(n): if s[i]!=t[i]: flag=True s=s[:i]+t[i]+s[i+1:] break if not flag: return False elif n+1==m: for i in range(n): if s[i]...
161One Edit Distance☢ 160Intersection of Two Linked ListsC 159Longest Substring with At Most Two Distinct Characters☢ 158Read N Characters Given Read4 II - Call multiple times☢ 157Read N Characters Given Read4☢ 156Binary Tree Upside Down☢ ...
161One Edit Distance♥Python1. Check the different position and conditions 2.Edit distance 163Missing Ranges♥PythonAdd -1 to lower for special case, then check if curr - prev >= 2 166Fraction to Recurring DecimalPython% and Hash to find duplicate ...