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：使用两个指针pointA代表0的位置，pointB代表2的位置。用一个for循环进行搜索，（1）遇到0的情况，直接存储到pointA的位置。（2）遇到2的情况，要进行判断，1）vpointB的位置是0吗，2）pointB位置是2吗，3）pointB的位置正好等于搜索的位置，对这三个情况进行处理。最后再用for[pointA,pointB]赋值为1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;
    }
};