[CC150][1.3] Given two strings, write a method to decide if one is a permutation of the other. (Consider case sensitivity and whitespace as significant)

permutation.java

import java.util.HashMap;

public class permutation {
	
	public static boolean isPermut(String a, String b) {
		if (a == null || b == null) throw new NullPointerException();
		if (a.length() != b.length()) return false;
		HashMap<Character, Integer> aMap = new HashMap<Character, Integer>();
		HashMap<Character, Integer> bMap = new HashMap<Character, Integer>();
		traverse(a, aMap);
		traverse(b, bMap);
		
		if(aMap.keySet().size() != bMap.keySet().size()) return false;
		
		for (Character key : aMap.keySet())
			if (!aMap.get(key).equals(bMap.get(key)))
				return false;
		return true;	
	}
	
	private static void traverse(String s, HashMap<Character, Integer> map) {
		Integer count;
		for (int i = 0; i < s.length(); i++) {
			count = map.get(s.charAt(i));
			if (count == null)
				map.put(s.charAt(i), 1);
			else map.put(s.charAt(i), count + 1);
		}
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		String a, b;
		a = "我为人人";
		b = "人人为我";
		System.out.println(permutation.isPermut(a, b) ? "Yes" : "No");
		
		a = "It's just a test right?";
		b = "It's just a right test?";
		System.out.println(permutation.isPermut(a, b) ? "Yes" : "No");
		
		a = "";
		b = " ";
		System.out.println(permutation.isPermut(a, b) ? "Yes" : "No");
		
		a = "Abcdef";
		b = "fedcba";
		System.out.println(permutation.isPermut(a, b) ? "Yes" : "No");
		
		a = "ありがとう";
		b = "うありがと";
		System.out.println(permutation.isPermut(a, b) ? "Yes" : "No");
	}

}