LeetCode: Single Number Ⅱ

/**
 * 
 */
package solution;

import java.util.Arrays;

/**
 * @author whh
 * 
 *         Given an array of integers, every element appears three times except
 *         for one. Find that single one.
 * 
 *         Note: Your algorithm should have a linear runtime complexity. Could
 *         you implement it without using extra memory?
 */
public class SingleNumber2 {

    /**
     * @param args
     */
    public static void main(String[] args) {
        int[] A1 = { 1 };
        int[] A2 = { 1, 2, 3, 4, 1, 2, 3, 1, 2, 3 };
        int[] A3 = { 1, 1, 2, 4, 2, 4, 3, 5, 3, 1, 2, 3, 4 };
        int[] A4 = { 1, 1 };
        int[] A5 = { 1, 1, 1, 2, 2, 2, 3, 3 };
        System.out.println(singleNumber(A1));
        System.out.println(singleNumber(A2));
        System.out.println(singleNumber(A3));
        System.out.println(singleNumber(A4));
        System.out.println(singleNumber(A5));
    }

    /**
     * 
     * @param A
     * @return
     */
    public static int singleNumber(int[] A) {
        Arrays.sort(A);
        int ret = 0;
        for (int i = 0; i < A.length; i = i + 3) {
            if (i == A.length - 1 || i == A.length - 2) {
                ret = A[i];
                break;
            } else if (!(A[i] == A[i + 1] && A[i] == A[i + 2])) {
                ret = A[i];
                break;
            }
        }
        return ret;
    }

}