Sort Colors - Leetcode

public class Solution {
    public void sortColors(int[] A) {
        int left=0, right=A.length-1, i=0;
        while(left<=right){
            if(A[left] == 0){
                swap(A,left,i);
               left++; i++;
            }else if(A[left] == 1){
              left++;
            }else{
                swap(A,left,right);
                right --;
            } 
        }
        
    }
    private void swap(int[]A, int l, int r){
        int x = A[l];
        A[l] = A[r];
        A[r] = x;
    }
}

分析：

1）走两遍的方法是：第一遍先记录每一个颜色的个数；第二遍根据个数一一分布和重新赋值颜色；不符合题目要求

2）走一遍的方法是：左指针和右指针分别代表2个颜色，第三个指针是游标，每遇到一个值进行判断：如果是0，则掉到左指针的位置；如果是1，则继续探索；如果是2，则调到右指针的位置。

3）看到网上还有一个方法是利用quick sort 的partition思想，这个思路可以分开N多颜色。 待续这个~

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?