Follow up:
A rather straight forward solution is a two-pass algorithm using counting sort.
First, iterate the array counting number of 0‘s, 1‘s, and 2‘s, then overwrite array with total number of 0‘s, then 1‘s and followed by 2‘s.

Could you come up with an one-pass algorithm using only constant space?

解题思路： 对数组进行排列，考的就是复杂度。使用两个指针pointA = 0，pointB = n - 1。进行一个for循环，（1）遇到数字0，A[pointA++] = 0；（2）遇到数字2，有三种情况1）pointB的位置是0。2）pointB位置是2,。3）pointB位置与for循环i的位置一样。对这几种情况进行处理就行。

class Solution {
public:
    void sortColors(int A[], int n) {
        int pointA = 0;
        int pointB = n-1;
        for (int i = 0; i <= pointB; i++) {
            if (A[i] == 0)
                A[pointA++] = 0;
            else if (A[i] == 2) {
                   if (A[pointB] == 0) {
                       A[pointA++] = 0 ;
                       A[pointB--] = 2;
                   } else if(A[pointB] == 2) {
                     if(i == pointB) {
                       pointB--;
                     } else {
                       while(A[pointB] == 2) {
                         pointB--;
                         if(i == pointB)
                         {
                           break;
                         }
                       }
                       if(A[pointB]== 0) {
                         A[pointA++] = 0 ;
                         A[pointB--] = 2;
                       } else {
                         A[pointB--] = 2;
                       }
                     }
                   } else A[pointB--] = 2;
            }
        }
        for(int i = pointA;i <= pointB;i++)
            A[i] = 1;
    }
};