動機

有點像287一直換的解法，但後面還可以擴充成two pointer去夾

Problem

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.

We will use the integers 0, 1, and 2 to represent the color red, white, and blue, respectively.

You must solve this problem without using the library's sort function.

 

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]

 

Constraints:

  • n == nums.length
  • 1 <= n <= 300
  • nums[i] is 0, 1, or 2.

 

Follow up: Could you come up with a one-pass algorithm using only constant extra space?

Sol

從最小的color開始，往pos的位置換

class Solution:
    def sortColors(self, nums: List[int]) -> None:
        pos,color = 0,0
        while pos < len(nums):
            for i in range(pos,len(nums)):
                if nums[i] == color:
                    nums[i], nums[pos] = nums[pos], nums[i]
                    pos += 1
            color += 1

上面只有看一邊，因為只有三個顏色，所以可以放兩個ptr一個看最小，一個看最大

但要注意，如果是從最大的換過來的，還要再檢查一次，因為最大的range中可能有比他小的像0或是1

class Solution:
    def sortColors(self, nums: List[int]) -> None:
        left,right,cur = 0,len(nums)-1,0
        while cur <= right:
            if nums[cur] == 0:
                nums[cur], nums[left] = nums[left], nums[cur]
                left += 1
            elif nums[cur] == 2:
                nums[cur], nums[right] = nums[right], nums[cur]
                right -= 1
                cur -= 1 # 再檢查一次
            cur += 1