242. Valid Anagram

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解法一：统计词频

class Solution {
public:
    bool isAnagram(string s, string t) {
        int word_s[26] = {0};
        int word_t[26] = {0};
        cout<

解法二：先排序，然后比较

class Solution {
public:
    bool isAnagram(string s, string t) {
        sort(s.begin(), s.end());
        sort(t.begin(), t.end());
        if(s == t){
            return true;
        }
        else{
            return false;
        }
    }
};