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) * ;
     int count[L], i, j, result = ;
     fill_n(&count[], L, );
     for (i = ; i < L; i++)
     {
         for (j = ; j < n; j++)
         {
             count[i] += ((A[j] >> i) & );
             count[i] %= ;
         }
     }

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