75. 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?

　这道题目给了我们一个颜色的array，让我们sort一下，按照0，1，2的顺序。根据题目要求，我们只能遍历array一次，可以用到two pointers来实现。设一个指针red 在开头，blue 在最后。想法就是，遇到红色0，就交换，把0放到最左边去；遇到蓝色2就交换，把2都放到最右边去，这样1就会被保留在最中间。需要注意的是，当把蓝色2交换完毕之后，需要i--， 停留 i 在原地一次，因为还需要继续检查 被2交换回来的数字。那当遇到红色0，交换完毕不需要停留i 的原因是， 交换回来的只可能是1，对于1，我们不需要做任何处理，直接过就可以。

 

 

关键词：Array

关键点：用two pointers，一头一尾放置红色和蓝色，保留白色在中间

class Solution {
public:
	void sortColors(vector<int>& nums) {
		int red = 0; 
		int blue = nums.size() - 1; 
		int temp; 


		for (int i = 0; i <= blue; ++i)
		{
			if (nums[i] == 0)
			{
				swap(nums[i], nums[red]);
				red++;
			}
			else if (nums[i] == 2)
			{
				swap(nums[i], nums[blue]);
				blue--;

				i--;
			}
		}
	}
	void swap(int & a, int & b)
	{
		int temp = a;
		a = b;
		b = temp;
	}
};