【Sort】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.

Follow up:
What if the inputs contain unicode characters? How would you adapt your solution to such case?

AC代码：

class Solution {public:    bool isAnagram(string s, string t) {        unordered_map<int,int> map;        unordered_map<int,int>::iterator it;        int len1=s.length();        int len2=t.length();        if(len1!=len2) return false;        for(int i=0;i<len1;i++){            map[s[i]-'a']++;            map[t[i]-'a']--;        }        for(it=map.begin();it!=map.end();it++)            if((it->second)!=0)                return false;        return true;    }};

这是人家的哈希表，优雅多了。。。哎。。。自己还是很菜啊，不熟悉C++的STL

class Solution {public:    bool isAnagram(string s, string t) {        if (s.length() != t.length()) return false;        int n = s.length();        unordered_map<char, int> counts;        for (int i = 0; i < n; i++) {            counts[s[i]]++;            counts[t[i]]--;        }        for (auto count : counts)            if (count.second) return false;        return true;    }};

事实上用数组就可以办到了：

class Solution {public:    bool isAnagram(string s, string t) {        if (s.length() != t.length()) return false;        int n = s.length();        int counts[26] = {0};        for (int i = 0; i < n; i++) {             counts[s[i] - 'a']++;            counts[t[i] - 'a']--;        }        for (int i = 0; i < 26; i++)            if (counts[i]) return false;        return true;    }};

然后当然最直接的方法是给两个数组排序，然后比较一下一不一样，啊哈哈哈。但是耗时肯定很长了=。=