Single Number II

题目：

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?

我的解法：

（1）算法思想：

本题是Single Number 的升级版（见博文《Single Number》）。先用快排算法将数组排序，然后遍历一次数组即可：对于i，如果A[i]=A[i+1]，则将i加上3，否则返回A[i].

（2）代码如下：

class Solution {
public:
int partition(int A[],int low,int high){
int pivot=A[low];
while(low=pivot&&high>low) --high;
A[low]=A[high];
while(A[low]<=pivot&&low