之前轉載過一篇STL的sort方法底層詳解的部落格：https://www.cnblogs.com/ygh1229/articles/9806398.html

但是我們在開發中會根據自己特定的應用，有新的排序需求，比如下面這道題，當只有0，1，2這三個數字時的排序，我們就可以自己寫定製版的排序演算法

描述

Given an array with n objects colored red, white or blue, sort them in-place 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.

Example:

Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

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 a one-pass algorithm using only constant space?

思路

我們定義 Low, Mid and High.

    1 0 2 2 1 0
    ^         ^
    L         H
    M

    Mid != 0 || 2
    Mid++

    1 0 2 2 1 0
    ^ ^       ^
    L M       H

    Mid == 0
    Swap Low and Mid
    Mid
 
++
    Low++

    0 1 2 2 1 0
      ^ ^     ^
      L M     H

    Mid == 2
    Swap High and Mid
    High--

    0 1 0 2 1 2
      ^ ^   ^
      L M   H

    Mid == 0
    Swap Low and Mid
    Mid++
    Low++

    0 0 1 2 1 2
        ^ ^ ^
        L M H

    Mid == 2
    Swap High and Mid
    High--

    0 0 1 1 2 2
        ^ ^
        L M
          H

    Mid <= High is our exit case

依照此思路，演算法解答如下

class Solution {
public:
    void sortColors(vector<int>& nums) 
    {
        int tmp = 0, low = 0, mid = 0, high = nums.size() - 1;
    
        while(mid <= high)
        {
            if(nums[mid] == 0)
            {
                tmp = nums[low];
                nums[low] = nums[mid];
                nums[mid] = tmp;
                low++;
                mid++;
            }
            else if(nums[mid] == 1)
            {
                mid++;
            }
            else if(nums[mid] == 2)
            {
                tmp = nums[high];
                nums[high] = nums[mid];
                nums[mid] = tmp;
                high--;
            }
        }
    }
};