LeetCode:Single Number(2)

问题描述：

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：将数组中的每一个元素的每一位相加，然后mod 3。得到的就是要求的数相应位上的元素。最后再将32个位组合，既得要求的数。

代码：

int Solution::singleNumber(int A[], int n){    int bits[32] = {0};    int sum;    for(int i = 0;i < 32;i++){        sum = 0;        for(int j = 0;j < n;j++){        if((A[j] >> i) & 1 != 0)            sum = sum + 1;        }        bits[i] = sum % 3;    }    int num = 0;    int x = 1;    for(int i = 0;i < 32;i++){        num = num + bits[i] * x;        x = x * 2;    }    return num;}