Single Number2

题目链接:http://oj.leetcode.com/problems/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?

解法:由于数组中所有元素都是整型,整型长度为32位,可以用一个数组count[32]记录所有元素按位出现1的总数,然后count[i]=count[i]%3,这样做可以将出现三次元素的每位

　　置为0,最后数组count[32]为出现一次元素的每位.

class Solution {
public:
    int singleNumber(int A[], int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        
    const int L = sizeof(int) * 8;
    int count[L], i, j, result = 0;
    fill_n(&count[0], L, 0);
    for (i = 0; i < L; i++)
    {
        for (j = 0; j < n; j++)
        {
            count[i] += ((A[j] >> i) & 1);
            count[i] %= 3;
        }
    }


    for (i = 0; i < L; i++)
    {
        result += (count[i] << i);
    }
    return result;
    }
};