Leetcode 75. Sort Colors

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问题描述

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?

解题思路

基本按照计数排序的思路，先遍历统计0、1、2的个数。根据个数重新对数组赋值。

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class Solution:
    def sortColors(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        count = [0, 0, 0]
        for i in nums:
            count[i] += 1
        print(count)
        print(nums[:count[i]])
        nums[:count[0]] = [0]*count[0]
        nums[count[0]:count[1]+count[0]] = [1]*count[1]
        nums[count[1]+count[0]:] = [2]*count[2]

进阶思路

上述做法貌似需要 \(O(n)\) 的额外存储空间，如果 \(O(1)\) 来做又会导致遍历两次。因此可以考虑类似快速排序的做法，两个指针分别指向头尾，一个指针从左到右移动。。将遇到的2移动到最右边，遇到的0移动到最左边。

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class Solution:
    def sortColors(self, nums: 'List[int]') -> 'None':
        """
        Do not return anything, modify nums in-place instead.
        """
        if len(nums)<=1:
            return

        l = 0
        r = len(nums) - 1
        i=0
        while i<=r:
            if nums[i] == 2 and i<r:
                nums[i], nums[r] = nums[r], nums[i]
                r-=1
            elif nums[i] == 0 and i>l:
                nums[i], nums[l] = nums[l], nums[i]
                l+=1
            else:
                i+=1
        return