【LeetCode】【定制版排序】Sort Colors

之前转载过一篇STL的sort方法底层详解的博客：https://www.cnblogs.com/ygh1229/articles/9806398.html

但是我们在开发中会根据自己特定的应用，有新的排序需求，比如下面这道题，当只有0，1，2这三个数字时的排序，我们就可以自己写定制版的排序算法

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?

思路一 标识位移动

我们定义 Low, Mid and High.

    ^         ^
    L         H
    M

    Mid !=  ||
    Mid++

    ^ ^       ^
    L M       H

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

      ^ ^     ^
      L M     H

    Mid ==
    Swap High and Mid
    High--

      ^ ^   ^
      L M   H

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

        ^ ^ ^
        L M H

    Mid ==
    Swap High and Mid
    High--

        ^ ^
        L M
          H

    Mid <= High is our exit case

依照此思路，算法解答如下

class Solution {
public:
    void sortColors(vector<int>& nums)
    {
        int tmp = , low = , mid = , high = nums.size() - ;

        while(mid <= high)
        {
            if(nums[mid] == )
            {
                tmp = nums[low];
                nums[low] = nums[mid];
                nums[mid] = tmp;
                low++;
                mid++;
            }
            else if(nums[mid] == )
            {
                mid++;
            }
            else if(nums[mid] == )
            {
                tmp = nums[high];
                nums[high] = nums[mid];
                nums[mid] = tmp;
                high--;
            }
        }
    }
};

思路三  优化

优化为只找序列中的0，2并且每次移位都要移彻底，调换位置后可能还会出现0，2，所以在while中位移完全结束后才继续向前执行。

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int low = , high = nums.size() - ;
        for(int i = ;i<=high;++i){
            while (nums[i]== && i<high) swap(nums[i], nums[high--]);
            while (nums[i]== && i>low) swap(nums[i], nums[low++]);
        }
    }
};