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?

要求遍历一遍且使用有限的空间，我想了很久都是需要两次遍历或是额外的空间，看了一个思路很简洁的代码，觉得不错分享给大家。但这个代码也有缺点，就是类似于插入排序，需要移动大量元素的位置，时间复杂度也是蛮高的。

class Solution:
    # @param A a list of integers
    # @return nothing, sort in place
    def sortColors(self, A):
        i=j=k=-1
        n=len(A)
        for m in range(n):
            if A[m]==0:
                k=k+1
                A[k]=2
                j=j+1
                A[j]=1
                i=i+1
                A[i]=0
            elif A[m]==1:
                k=k+1
                A[k]=2
                j=j+1
                A[j]=1
            else:
                k=k+1
                A[k]=2