Leetcode Valid Anagram高效解法

// Author : yqtao// Date   : 2016-6.19// Email  ：[e-mail]/************************************************************************************ Given two strings s and t, write a function to determine if t is an anagram of s.*                                              即判断t是不是s字母交换顺序组成的字母* For example,* s = "anagram", t = "nagaram", return true.* s = "rat", t = "car", return false.** Note:* You may assume the string contains only lowercase alphabets.***********************************************************************************/#include "stdafx.h"#include <iostream>#include <string>#include <vector>#include <algorithm>using namespace std;//很容易理解的方法，即排序//但是时间复杂度很高bool isAnagranm(string &s, string &t){    sort(s.begin(), s.end());    sort(t.begin(), t.end());    if (s == t)        return true;    else        return false;}//第二种方法，使用哈希表bool isAnagranm1(string &s, string &t){    int map[26] = { 0 };    for (int i = 0; i < s.size(); i++)        map[s[i] - 'a']++;//将字母存放在map的矩阵中    for (int i = 0; i < t.size(); i++)        map[t[i] - 'a']--;    for (int i = 0; i < sizeof(map) / sizeof(map[0]); i++)        if (map[i] != 0)            return false;    return true;//整个map矩阵全为0，则为true}int main(){    string s = "nagaram";    string p = "nagaram";    cout << boolalpha << isAnagranm(s, p) << endl;//80ms    cout << boolalpha << isAnagranm1(s, p) << endl;//12ms}