LeetCode刷题笔录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?

这题一开始理解有误，以为另外那个数字可以出现N次，只要N不是3.后来发现N只能是1.

方法是把所有数字的32个bit分别加起来。如果所有数字都出现三次，那么32位每一位上最后的和都是3的倍数应该。因此把32个结果%3得到的余数就是单个的数字。

public class Solution {    public int singleNumber(int[] A) {        int[] bits = new int[32];        Arrays.fill(bits, 0);        for(int i = 0; i < A.length; i++){            int mask = 1;            for(int j = 0; j < 32; j++){                int shifted = A[i] >> j;                if(shifted == 0)                    break;                else                    bits[j] += shifted & mask;            }        }                int result = 0;        for(int i = 0; i < 32; i++){            result += (bits[i] % 3) << i;        }        return result;    }}

另外这里有个很神的blog，里面有种高端的方法，还没深入理解，先记下来。