Single Number II

Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

Input: [2,2,3,2]
Output: 3

Example 2:

Input: [0,1,0,1,0,1,99]
Output: 99

就是 Int 32位，因为每个数字appear 3次，所以，每个位要么为3K+1,或者为3K.

取决于只出现一次的位是0，还是1.

这样思路就是取每一位相加，然后%3, 这样下来就是那个出现1次的数的位数，然后移动回去，加起来就是出现一次的数。

class Solution {
    public int singleNumber(int[] nums) {
        int single = 0;
        for(int i = 0; i < 32; i++) {
            int bit = 0;
            for(int j = 0; j < nums.length; j++) {
                bit += (nums[j] >> i) & 1;
            }
            bit = bit % 3;
            single += (bit << i);
        }
        return single;
    }
}