Sort Colors

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.

click to show follow up.

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?

计数排序    void sortColors(int A[], int n)     {        int red_count = 0;        int white_count = 0;      //  int blue_count = 0;                 for(int i = 0;i < n;i++)        {            if(A[i] == 0)            {                red_count++;            }            else if(A[i] == 1)            {                white_count++;            }        }                for(int i = 0;i< n;i++)        {            if(red_count > 0)            {                A[i] = 0;                red_count--;            }            else if(white_count > 0)            {                A[i] = 1;                white_count--;            }            else            {                A[i] = 2;            }        }    }

思路2.中间元素，分派给俩边！

也可参考：RGB排序 更不易出错！！

//俩边扫描，（需至少，3个索引，也可4个人。俩个记录左右，一个往前走的）//将中间的数据与俩边交换void sortColors(int A[], int n) {    int pos = 0;    int low = 0;    int high = n - 1;        while( pos <= high)    {        if( A[pos] == 0)        {            if(A[low] != 0)            {                swap(A[low],A[pos]);            }            low ++;            pos++;//前面扫描过直接走，交换过来的不可能为2        }        else if(A[pos] == 2)        {             if( A[high] != 2)             {                 swap(A[high],A[pos]);             }             high -- ; //此处pos不动，关键                  }        else        {            pos++;        }            }    }