problem：
Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue.
Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.
Note:
You are not suppose to use the library‘s sort function for this problem.
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?
thinking：
（1）最简单的是使用计数法，计算0、1、2 的个数，最后再填充
（2）最优算法是 快速排序的变形：一次遍历完成排序
code：
class Solution {
public:
void sortColors(int A[], int n) {
int i = 0; // 0 pointer
int j = n - 1; // 1 pointer
int k = n - 1; // 2 pointer
while (i <= j)
{
if (A[i] == 2)
{
int t = A[k];
A[k] = A[i];
A[i] = t;
k--;
if (k < j)
j = k;
}
else if (A[i] == 1)
{
int t = A[j];
A[j] = A[i];
A[i] = t;
j--;
}
else
i++;
}
}
};原文地址：http://blog.csdn.net/hustyangju/article/details/44961465
