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…

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.…

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…

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>…

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…

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…

题意:取出字符串Str里的两个串S,T,问对应位置的的字符在否有一一映射关系. hash:对于每个字符s=‘a’-‘z’,我们任意找一个i,满足Si==s,(代码里用lower_bound在区间找到最小的位置i)其对应的字符为Ti==t.然后我们判断这段区间里s的hash值是否等于t的hash值.不难证明26个字母都满足时对应hash相同时,才有一一映射关系.(但是不明白我的自然溢出的hash版本为什么错了,但是mod(1e9+7)的版本对了? #include<bits/stdc++.h>…

205. 同构字符串 205. Isomorphic Strings…

Isomorphic Strings 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 or…

Question 205. Isomorphic Strings Solution 题目大意:判断两个字符串是否具有相同的结构 思路:构造一个map,存储每个字符的差,遍历字符串,判断两个两个字符串中相同位置字符的差是否相同 Java实现: public boolean isIsomorphic(String s, String t) { Map<String, Integer> map = new HashMap<>(); for (int i=0; i<s.length(…

Isomorphic Strings 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 preservin…

Codeforces Educational Codeforces Round 44 (Rated for Div. 2) F. Isomorphic Strings 题目连接: http://codeforces.com/contest/985/problem/F Description You are given a string s of length n consisting of lowercase English letters. For two given strings s an…

F - Isomorphic Strings 思路:字符串hash 对于每一个字母单独hash 对于一段区间,求出每个字母的hash值,然后排序,如果能匹配上,就说明在这段区间存在字母间的一一映射 代码: #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair…

F - Isomorphic Strings 题目大意:给你一个长度为n 由小写字母组成的字符串,有m个询问, 每个询问给你两个区间, 问你xi,yi能不能形成映射关系. 思路:这个题意好难懂啊...  字符串哈希, 将26个字符分开来hash, 那么check就变成啦, 区间内对应的26个字符的hash值是否一致. 即如果 a -> b  那么区间1内a的hash值等于区间2内b的hash值. #include<bits/stdc++.h> #define LL long long #…

LeetCode--Add Strings Question Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2. Note: The length of both num1 and num2 is < 5100. Both num1 and num2 contains only digits 0-9. Both num1 and num2 doe…

205. Isomorphic Strings 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 character must be replaced with another character while preserv…

记得曾经一次面试时,面试官给我电脑,让我现场写个算法,判断一个字符串是不是对称字符串.我当时用了几分钟写了一个很简单的代码. 这里说的对称字符串是指字符串的左边和右边字符顺序相反,如"abba",单个字符串暂且算非对称字符串,虽然有字符串看起来是对称的如"A."H"."O"."中"."人"...,严格来说,也是不对称的,把它们放大时,线条的粗细是不一样的. static bool IsSymmet…

 题目…

import java.util.Scanner; /** *        判断一个字符串是否是对称字符串 */ public class StringDemo { public static void main(String[] args) { // 使用键盘录入获取字符串 Scanner sc = new Scanner(System.in); // 友情提示 System.out.println("请输入一个字符串"); // 接收录入的字符串 String str = sc.…

Leetcode(8)字符串转换整数 [题目表述]: 请你来实现一个 atoi 函数,使其能将字符串转换成整数. 首先,该函数会根据需要丢弃无用的开头空格字符,直到寻找到第一个非空格的字符为止. 当我们寻找到的第一个非空字符为正或者负号时,则将该符号与之后面尽可能多的连续数字组合起来,作为该整数的正负号:假如第一个非空字符是数字,则直接将其与之后连续的数字字符组合起来,形成整数. 该字符串除了有效的整数部分之后也可能会存在多余的字符,这些字符可以被忽略,它们对于函数不应该造成影响. 注意:假如该…

F. Isomorphic Strings time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output You are given a string s of length n consisting of lowercase English letters. For two given strings s and t, say S is…

问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/3770 访问. 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. 输入: s = "egg", t = "add" 输出: true 输入: s =…

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.…

翻译 给定两个字符串s和t,决定它们是否是同构的. 假设s中的元素被替换能够得到t,那么称这两个字符串是同构的. 在用一个字符串的元素替换还有一个字符串的元素的过程中.所有字符的顺序必须保留. 没有两个字符能够被映射到同样的字符,但字符能够映射到该字符本身. 比如, 给定"egg","add",返回真. 给定"foo"."bar",返回假. 给定"paper","title",返回真.…

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…

判断两个字符串是否同构 hs,ht就是每个字符出现的顺序 "egg" 与"add"的数字都是122 "foo"是122, 而"bar"是123 class Solution { public: bool isIsomorphic(string s, string t) { ] = {}; ] = {}; if (s.size() != t.size()) return false; ; ;i<s.size(); ++i)…