Sort Colors -- leetcode

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?

算法一：红白蓝三指针从前向后移动

这个算法理解可以借用刷墙。

假设有x个红，y个白，z个蓝

先全部刷成蓝（x+y+z)。

再从头刷(x+y)个成白。

再从头刷x个成红。

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

算法参考自：
https://leetcode.com/discuss/1827/anyone-with-one-pass-and-constant-space-solution

算法二：红白二指针从前向后， 蓝指针从后往前移动

class Solution {
public:
    void sortColors(int A[], int n) {
        int red = -1, white = 0, blue = n;
        while (white < blue) {
            if (A[white] == 1) {
                ++white;
            }
            else if (A[white] == 0) {
                A[white++] = 1;
                A[++red] = 0;
            }
            else {
                A[white] = A[--blue];
                A[blue] = 2;
            }
        }
    }
};