Given an array nums 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.
Follow up: • Could you solve this problem without using the library’s sort function? • Could you come up with a one-pass algorithm using only O(1) constant space?
Constraints: • n == nums.length • 1 <= n <= 300 • nums[i] is 0, 1, or 2.
Example 1: Input: nums = [2,0,2,1,1,0] Output: [0,0,1,1,2,2]
Example 2: Input: nums = [2,0,1] Output: [0,1,2]
Example 3: Input: nums = [0] Output: [0]
Example 4: Input: nums = [1] Output: [1]
解题思路为利用左右双指针。left指针指向要替换成0的位置，right指针指向要替换成2的位置。遍历数组， 当遍历元素为0时，则比较当前位置是否大于left指针位置，如果相等，说明该位置之前都为0，不需要操作，left指针后移一位继续遍历；如果当前位置大于left指针位置，则将当前元素与指针位置元素进行交换，并left指针后移一位，继续遍历下一元素。 当遍历元素为2时，当前位置与right指针位置比较，如果right指针指向为2，则right指针前移一位。如果right指针指向不为2，那么将当前元素与指针位置元素进行交换，并且right指针前移一位，继续判断交换的i元素。 当遍历元素为1时，继续遍历。 当遍历的位置等于right指针位置时，即结束遍历。
实现代码如下：
class Solution: def sortColors(self, nums: List[int]) -> None: """ Do not return anything, modify nums in-place instead. """ left = 0 right = len(nums) - 1 i = 0 while i <= right : if nums[i] == 0: if i > left: nums[left], nums[i] = nums[i], nums[left] i += 1 left += 1 elif nums[i] == 2: if nums[right] != 2: nums[right], nums[i] = nums[i], nums[right] right -= 1 else: i += 1