# 242.Valid Anagram

**242.Valid Anagram**

难度:Easy

> 给定两个字符串 s 和 t ，编写一个函数来判断 t 是否是 s 的一个字母异位词。

```
示例 1:

输入: s = "anagram", t = "nagaram"
输出: true
示例 2:

输入: s = "rat", t = "car"
输出: false
说明:
你可以假设字符串只包含小写字母。
```

进阶: 如果输入字符串包含 unicode 字符怎么办？你能否调整你的解法来应对这种情况？

可以使用map映射比较，由于一共24个小写字母，也可以直接使用数组比较。

```
class Solution {
public:
    bool isAnagram(string s, string t) {
       int s1[26]={0};
        int t1[26]={0};
        for(auto c:s)
            s1[c-'a']++;
        
        for(auto c:t)
            t1[c-'a']++;
        for(int i=0;i<26;i++)
            if(s1[i]!=t1[i]) return false;
        return true;
    }
};
```

> 执行用时 : 12 ms, 在Valid Anagram的C++提交中击败了97.08% 的用户\
> 内存消耗 : 9.5 MB, 在Valid Anagram的C++提交中击败了8.77% 的用户


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