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?

思路：就是用array来记录出现次数，最后再减去第二个string的出现次数，最后应该是0. othewise false;

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

Follow up question: 如果是Unicode，那就用hashmap。可以用一个hashmap，一个string只负责+1，一个string只负责-1

最后的hashmap每个key应该得到0，否则false；

class Solution {
    public boolean isAnagram(String s, String t) {
        if(s == null && t == null) return true;
        if(s == null || t == null || s.length() != t.length()) return false;
        
        HashMap<Character, Integer> smap = new HashMap<Character, Integer>();
        
        for(int i=0; i<s.length(); i++){
            Character sc = s.charAt(i);
            Character tc = t.charAt(i);
            if(smap.get(sc)!=null){
                smap.put(sc, smap.get(sc)+1);
            } else {
                smap.put(sc,1);
            }
            
            if(smap.get(tc)!=null){
                smap.put(tc, smap.get(tc)-1);
            } else {
                smap.put(tc,-1);
            }
        }
       
        
        for(Character sc: smap.keySet()){
           if(smap.get(sc)!=0){
               return false;
           }
        }
        return true;
    }
}