LeetCode-Single Number II

题目：https://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?
分析：创建一个长度为sizeof(int) 的数组count[sizeof(int)]，count[i] 表示在在i 位出现的1 的次数。如果count[i] 是3 的整数倍，则忽略；否则就把该位取出来组成答案。
源码：Java版本

算法分析：时间复杂度O(n)，空间复杂度O(1)

public class Solution {
    public int singleNumber(int[] A) {
         int[] count=new int[Integer.SIZE];
		 int result=0;
		 for(int i=0;i<A.length;i++) {
			 for(int j=0;j<Integer.SIZE;j++) {
				 if(((1<<j)&A[i])!=0) {
					 count[j]++;
					 count[j]%=3;
				 }
			 }
		 }
		 for(int i=0;i<Integer.SIZE;i++) {
			 result+=count[i]<<i;
		 }
		 return result;
    }
}