****(leetcode) 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?

思路：之前single number那个题是 every element appears twice except for one，那个是异或即可

出现三次的话，把每个数看做二进制，每个位置上的0/1相加，然后对三取模，所的得即是最后结果的那位数字。

class Solution {
public:
    int singleNumber(int A[], int n) {
        int bin[32] = {0};
        int ans = 0;
        for(int i = 0; i < 32 ; i++){
            for(int j = 0 ; j < n ; j++){
                bin[i]+=(A[j]>>i)&1; //判断第i位 是否为1 
            }
            ans|=(bin[i]%3)<<i; //<<即相当于乘以加权数，*2，*2^2， *2^3 ...
        }
        return ans;
    }
};