Single Number II

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 int singleNumber(int[] A) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        Hashtable<Integer, Integer> table = new Hashtable<Integer, Integer>();
        for(int i = 0; i < A.length; i++){
            if(table.containsKey(A[i]))
                table.put(A[i], table.get(A[i]) + 1);
            else
                table.put(A[i], 1);
        }
        for(int i = 0; i < A.length; i++){
            if(table.get(A[i]) != 3)
                return A[i];
        }
        return 0;
    }

Hash uses extra space.

Bit operation does not need extra space.

	public int singleNumber2(int[] A){
		int ones = 0, twos = 0;
		int commonBitMask = 0;
		for(int i = 0; i < A.length; i++){
			twos |= (ones & A[i]);
			ones ^= A[i];
			commonBitMask = ~(ones & twos);
			ones &= commonBitMask;
			twos &= commonBitMask;
		}
		return ones;
	}

非常牛逼的解法

	public int singleNumber3(int A[]) {
		int[] count = new int[32];
		int result = 0;
		for (int i = 0; i < 32; ++i) {
			count[i] = 0;
			for (int j = 0; j < A.length; ++j) {
				if (((A[j] >> i) & 1) != 0)
					count[i] = (count[i] + 1) % 3;
			}
			result |= (count[i] << i);
		}
		return result;
	}