242. Valid Anagram*

242. Valid Anagram*

​​https://leetcode.com/problems/valid-anagram/​​

题目描述

Given two strings ​​s​​​ and ​​t​​​ , write a function to determine if ​​t​​​ is an anagram of ​​s​​.

Example 1:

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

Example 2:

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

Note:
You may assume the string contains only lowercase alphabets.

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

C++ 实现 0

20210323 更新: 使用哈希表, 注意最后还要保证哈希表为空.

class Solution {
public:
    bool isAnagram(string s, string t) {
        unordered_map<char, int> record;
        for (auto &c : s) record[c] ++;
        for (auto &c : t) {
            if (!record.count(c)) return false;
            record[c] --;
            if (record[c] == 0) record.erase(c);
        }
        return record.empty();
    }
};

C++ 实现 1

Anagram 应该是说 ​​t​​​ 是由 ​​s​​​ 中的字符串任意排列形成的. 要判断 ​​t​​​ 是否为 ​​s​​​ 的 Anagram, 首先要判断它们大小是否相等, 其次判断 ​​t​​​ 中的每种字符的个数是否等于 ​​s​​ 中每种字符的个数.

class Solution {
public:
    bool isAnagram(string s, string t) {
        if (s.size() != t.size()) return false;
        unordered_map<char, int> A, B;
        for (auto &c : s) A[c] ++;
        for (auto &c : t) B[c] ++;
        for (auto &p : A) {
            if (!B.count(p.first) || B[p.first] != p.second) return false;
        }
        return true;
    }
};

C++ 实现 2

由于题目中已经说明 ​​s​​​ 和 ​​t​​ 中只包含小写字母, 那么也可以用数组来表示哈希表.

class Solution {
public:
    bool isAnagram(string s, string t) {
        if (s.size() != t.size()) return false;
        int record[256] = {0};
        for (int i = 0; i < s.size(); ++i) {
            record[s[i]] ++;
            record[t[i]] --;
        }
        for (int i = 0; i < 256; ++i)
            if (record[i] != 0)
                return false;
        return true;
    }
};