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?

使用set集合排序 Runtime: 236 ms

class Solution {public:    bool isAnagram(string s, string t) {        multiset<char> set1;        multiset<char> set2;        for(auto it=s.begin();it!=s.end();it++)            set1.insert(*it);        for(auto it=t.begin();it!=t.end();it++)            set2.insert(*it);            if(set1!=set2)            return false;        return true;    }};

使用sort算法  Runtime: 80 ms

class Solution {public:    bool isAnagram(string s, string t) {        sort(s.begin(),s.end());        sort(t.begin(),t.end());        return s==t;    }};

使用hash table  unordered_map   Runtime: 36 ms

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

使用数组代替unordered_map获得更短的时间 Runtime: 12 ms

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