LeetCode 솔루션

[7/28] 242. Valid Anagram

학부유학생 쪽지보내기 메일보내기 자기소개 아이디로 검색 전체게시물 작성
작성일 2022.07.27 21:50

6,931 조회
4 댓글
0 추천
0 비추천
목록 게시글 쓰기

Easy

5889232Add to ListShare

Given two strings s and t, return true if t is an anagram of s, and false otherwise.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Example 1:

Input: s = "anagram", t = "nagaram"
Output: true

Example 2:

Input: s = "rat", t = "car"
Output: false

Constraints:

1 <= s.length, t.length <= 5 * 104
s and t consist of lowercase English letters.

Follow up: What if the inputs contain Unicode characters? How would you adapt your solution to such a case?

Accepted

1,473,092

Submissions

2,373,483

#Amazon, #Bloomberg, #Spotify, #Apple, #Facebook, #Microsoft, #Affirm, #Google, #Yandex, #Yahoo, #Tesla

링크

https://leetcode.com/problems/valid-anagram/ 2842 회 연결

익명
작성일 2022.07.27 21:51

Runtime: 73 ms, faster than 56.26% of Python3 online submissions for Valid Anagram.
Memory Usage: 14.4 MB, less than 96.92% of Python3 online submissions for Valid Anagram.

import collections
class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        return collections.Counter(s) == collections.Counter(t)

익명
작성일 2022.08.18 11:40

Runtime: 23 ms, faster than 22.74% of C++ online submissions for Valid Anagram.
Memory Usage: 7.4 MB, less than 46.64% of C++ online submissions for Valid Anagram.

public:
    bool isAnagram(string s, string t) {
        map<char, int> m;
        for(auto c : s)
            m[c]++;
        
        for(auto c : t)
        {
            if(m.find(c) == m.end())
                return false;
            m[c]--;
        }
        
        for(auto itr = m.begin(); itr != m.end(); itr++)
            if(itr->second != 0)
                return false;
        return true;
    }
};

익명
작성일 2022.08.18 11:43

Runtime: 16 ms, faster than 48.54% of C++ online submissions for Valid Anagram.
Memory Usage: 7.3 MB, less than 46.64% of C++ online submissions for Valid Anagram.

public:
    bool isAnagram(string s, string t) {
        sort(s.begin(), s.end() );
        sort(t.begin(), t.end() );
        
        if(s.size() != t.size())
            return false;
        for(int i = 0; i < s.size(); i++)
            if(s[i] != t[i])
                return false;
        
        
        return true;
    }
};

익명
작성일 2022.08.18 11:52

Runtime: 12 ms, faster than 64.75% of C++ online submissions for Valid Anagram.
Memory Usage: 7.3 MB, less than 78.25% of C++ online submissions for Valid Anagram.

class Solution {
public:
    bool isAnagram(string s, string t) {
        vector<int> map(26,0);
        
        for(char c : s)
            map[c-'a']++;
        for(char c : t)
            map[c-'a']--;
        
        for(auto it = map.begin(); it != map.end(); it++)
            if(*it != 0)
                return false;
        
        return true;
    }
};