题目：

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.

思路：

统计第一个字符串的每一个字符出现的次数，再统计第二个字符串的每一个字符出现的次数。然后比較，假设都同样则返回true。

假设不同返回false。

代码1：

class Solution {
public:bool isAnagram(string s, string t) {int len1 = s.size();int len2 = t.size();if(len1 != len2)return false;int map[26] = {0};//因为题目确定每一个字符都是小写字母。因此仅仅须要申请26个空间就可以for(int i = 0 ; i < len1 ; i++){map[s[i]-'a']++;}for(int i = 0 ; i < len1 ; i++){map[t[i]-'a']--;if(map[t[i]-'a'] < 0)return false;}return true;}
};

代码2：

class Solution {
public:bool isAnagram(string s, string t) {int len1 = s.size();int len2 = t.size();if(len1 != len2)return false;int map[26] = {0};//因为题目确定每一个字符都是小写字母，因此仅仅须要申请26个空间就可以for(int i = 0 ; i < len1 ; i++){map[s[i]-'a']++;map[t[i]-'a']--;}for(int i = 0 ; i < len1 ; i++){if(map[t[i]-'a'] != 0)return false;}return true;}
};

转载于:https://www.cnblogs.com/gavanwanggw/p/6878087.html

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LeetCode OJ 之 Valid Anagram

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代码2：

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