Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the characters in s can be replaced to get t. All occurrences of a character must be replaced with another character while preserving the order of characters.…

Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the characters in s can be replaced to get t. All occurrences of a character must be replaced with another character while preserving the order of characters.…

哈希表可以用ASCII码数组来实现,可以更快 public boolean isIsomorphic(String s, String t) { /* 思路是记录下每个字符出现的位置,当有重复时,检查另外一个map是不是也是对应位置重复 */ if (s.length()!=t.length()) { return false; } Map<Character,Integer> map1 = new HashMap<>(); Map<Character,Integer>…

给定两个字符串 s 和 t,判断它们是否是同构的.如果 s 中的字符可以被替换最终变成 t ,则两个字符串是同构的.所有出现的字符都必须用另一个字符替换,同时保留字符的顺序.两个字符不能映射到同一个字符上,但字符可以映射自己本身.例如,给定 "egg", "add", 返回 true.给定 "foo", "bar", 返回 false.给定 "paper", "title", 返回 tr…

题意:如果两个字符串是对称的,就返回true.对称就是将串1中的同一字符都一起换掉,可以换成同串2一样的. 思路:ASCII码表哈希就行了.需要扫3次字符串,共3*n的计算量.复杂度O(n).从串左开始扫,若字符没有出现过,则赋予其一个特定编号,在哈希表中记录,并将该字符改成编号.对串2同样处理.其实扫2次就行了,但是为了短码. class Solution { public: void cal(string &s) { ]={}, cnt=; ; i<s.size(); i++) { if…

题目: 给定两个字符串 s 和 *t*,判断它们是否是同构的. 如果 s 中的字符可以被替换得到 *t* ,那么这两个字符串是同构的. 所有出现的字符都必须用另一个字符替换,同时保留字符的顺序.两个字符不能映射到同一个字符上,但字符可以映射自己本身. Given two strings s* and t*, determine if they are isomorphic. Two strings are isomorphic if the characters in s* can be rep…

Isomorphic Strings Total Accepted: 30898 Total Submissions: 120944 Difficulty: Easy Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the characters in s can be replaced to get t. All occurrences of a characte…

Given two strings A and B of lowercase letters, return true if and only if we can swap two letters in A so that the result equals B. Example 1: Input: A = "ab", B = "ba" Output: true Example 2: Input: A = "ab", B = "ab&q…

问题描述: Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the characters in s can be replaced to get t. All occurrences of a character must be replaced with another character while preserving the order of charac…

题面戳我 Solution 我们按照每个字母出现的位置进行\(hash\),比如我们记录\(a\)的位置:我们就可以把位置表示为\(0101000111\)这种形式,然后进行字符串\(hash\) 每次查询时,我们就把两个子串的每个字母的\(hash\)值,取出来,判断能否一一对应即可 为啥我的常数那么大,2700ms Code //It is coded by ning_mew on 7.23 #include<bits/stdc++.h> #define LL long long usin…