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?

创建一个长度为32的数组a，a[i]表示所有数字在i位出现的次数。
假如a[i]是3的整数倍，则忽略；否则就把该位取出来组成答案。
空间复杂度O(1).

int sol1(int A[], int n)
{
    int count[32];
    int result = 0;
    for (int i = 0;i < 32;++i) {
        count[i] = 0;
        for (int j = 0;j < n;++j) {
            if ((A[j] >> i) & 1) count[i] = (count[i] + 1) % 3;
        }
        result |= (count[i] << i);
    }
    return result;
}

另一个方法，同样的原理，用3个整数来表示INT的各位的出现次数情况，one表示出现
了1次，two表示出现了2次。当出现3次的时候该位清零。最后答案就是one的值。

int sol2(int A[], int n)
{
    int one = 0, two = 0, three = 0;
    for (int i = 0;i < n;++i) {
        two |= one & A[i];
        one ^= A[i];
        three = one & two;
        one &= ~three;
        two &= ~three;
    }
    return one;
}

public class Solution {
    public int singleNumber(int[] A) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int count[] = new int[32];
        int result = 0;
        for (int i = 0; i < 32; i ++){
            for (int j = 0; j < A.length; j ++) {
                if (((A[j] >> i) & 1) == 1) 
                    count[i] = (count[i] + 1) % 3;
            }
            result |= (count[i] << i);
        }
        return result;
    }
}