*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.

解法一： sort！

public class Solution {
        public boolean isAnagram(String s, String t) {
            char[] schar = s.toCharArray();
            char[] tchar = t.toCharArray();
            Arrays.sort(schar);
            Arrays.sort(tchar);
            s = new String(schar);
            t = new String(tchar);
            return s.equals(t);
        }
    }

 解法二：hashmap

import java.util.*;
public class Solution {
    public boolean isAnagram(String s, String t) {
        if(s.length()!=t.length()) return false;
        
        Hashtable<Character, Integer> 
            theTableS = new Hashtable<Character, Integer>(),
            theTableT = new Hashtable<Character, Integer>();
        for (char i:s.toCharArray()) {
            if (theTableS.containsKey(i))
                theTableS.put(i,theTableS.get(i)+1);
            else theTableS.put(i,1);
        }
        for (char i:t.toCharArray()) {
            if (theTableT.containsKey(i)) 
                theTableT.put(i,theTableT.get(i)+1);
            else theTableT.put(i,1);
        }
        return theTableT.equals(theTableS);
    }
}

 解法三：O(n) time, O(1) space

public class Solution {
    public boolean isAnagram(String s, String t) {
        int[] alphabet = new int[26];
        for (int i = 0; i < s.length(); i++) alphabet[s.charAt(i) - 'a']++;
        for (int i = 0; i < t.length(); i++) alphabet[t.charAt(i) - 'a']--;
        for (int i : alphabet) if (i != 0) return false;
        return true;
    }
}