242. Valid Anagram

/* 242. Valid Anagram Given two strings s and t, write a function to determine if t is an anagram of s. For example, s = "anagram", t = "nagaram", return true. s = "rat", t = "car", return false. Note: You may assume the string contains only lowercase alphabets. 题目大意：判断t是不是s中字符乱序生成的字符串 解题思路： 方法1：对s和t排序，判断排序后的字符串是不是相等 方法2：打表，维护两个数组hashS和hashT,把s和t中的字符看成ASCII码，对每个出现的字母ch, 把对应的数组hash[ch]值+1。判断时比较hashS和hashT这两个数组是否相等。 */

#include <iostream>
#include <unordered_set>
#include <algorithm>
#include <string>

using namespace std;

class Solution {
public:
    bool isAnagram(string s, string t) 
    {
        int len1 = s.size();
        int len2 = t.size();
        sort(s.begin(), s.end());
        sort(t.begin(), t.end());

        return s == t;
    }

    bool isAnagram1(string s, string t)
    {
        int len1 = s.size();
        int len2 = t.size();
        if (len1 != len2)
            return false;
        vector<int> hashS(128, 0);
        vector<int> hashT(128, 0);

        for (int i = 0; i < len1; ++i)
        {
            hashS[s[i]]++;
            hashT[t[i]]++;
        }
        //由于只可能是小写字母，可以不用判断整个数组
        for (int i = 'a'; i <= 'z'; ++i)
        {
            if (hashS[i] != hashT[i])
                return false;
        }

        return true;
    } 
};

void test()
{
    Solution sol;
    string s1{ "aabcc" };
    string t1{ "abcac" };
    string s2{ "abc" };
    string t2{"bad"};
    cout << sol.isAnagram1(s1, t1) << endl;
    cout << sol.isAnagram(s2, t2) << endl;
}

int main()
{
    test();

    return 0;
}