[LeetCode] Valid Anagram

https://leetcode.com/problems/valid-anagram/?tab=Description

思路1：排序

Time complexity: O(nlogn)
Space complexity: O(1)

class Solution {
public:
    bool isAnagram(string s, string t) {
        sort(s.begin(), s.end());
        sort(t.begin(), t.end());
        return s == t;
        
        /* 千万不要像下面这么写！！！
        if (s == t) return true;
        return false;
        */
    }
};

思路2：Hash Table

Time complexity: O(n)
Space complexity: O(所有字符种类数量)，由于ASCII字符集一共有256种字符编码，所以可以视为常量

class Solution {
public:
    bool isAnagram(string s, string t) {
        if (s.size() != t.size()) {
            return false;
        }

        vector<int> count(256, 0);
        for (char c : s) {
            count[(int) c]++;
        }
        for (char c : t ) {
            count[(int) c]--;
            if (count[(int) c] < 0) {
                return false;
            }
        }
        return true;
    }
};