[LeetCode] (medium) 75. Sort Colors

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本文深入探讨了LeetCode上的一道经典题目——三色排序，即红、白、蓝三色球排序问题。通过分析两种不同的算法实现，一种是利用三个指针进行原地排序，另一种是使用计数排序的思想，但不使用额外的空间。文章详细解释了每种方法的步骤，并对比了它们的效率。

https://leetcode.com/problems/sort-colors/

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?

归根结底是利用了数组中元素种类已知（只有三种）这一特点，利用交换来代替整段平移。

参考：https://www.cnblogs.com/ganganloveu/p/3703746.html

解法一：

class Solution {
public:
    void sortColors(vector<int>& nums) {
        static int fast_io = []() { std::ios::sync_with_stdio(false); cin.tie(nullptr); return 0; }();
        int right = nums.size()-1;
        int left = 0;
        int cur = 0;
        while(cur <= right){
            if(nums[cur] == 1){
                ++cur;
            }else if(nums[cur] == 0){
                nums[cur] = nums[left];
                nums[left] = 0;
                ++left;
                ++cur;
                //cur = max(cur, left);
            }else if(nums[cur] == 2){
                nums[cur] = nums[right];
                nums[right] = 2;
                --right;
            }
        }
    }
};

解法二：

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int zero = 0;
        int one = 0;
        int two = 0;
        for(int i = 0; i < nums.size(); ++i){
            if(nums[i] == 2){
                nums[two] = 2;
                ++two;
            }else if(nums[i] == 1){
                nums[two] = 2;
                //nums[i] = 2;
                nums[one] = 1;
                ++two;
                ++one;
            }else{
                nums[two] = 2;
                //nums[i] = 2;
                nums[one] = 1;
                nums[zero] = 0;
                ++two;
                ++one;
                ++zero;
            }
        }
    }
};

*解法二中，利用i来代替two反而会变慢，估计是编译优化里的问题，我不知道是为什么。