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;

}
};