Leetcode.75 | Sort Colors
阿新 • • 發佈:2020-07-27
Leetcode.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:
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?
Solution
題目相當於0、1、2排序,而且是有重複資料的排序
解法一:計數排序
class Solution: def sort_colors_violent(self, nums: list) -> None: pass def sort_colors_count_sort(self, nums: list) -> None: """ Do not return anything, modify nums in-place instead. """ # 定義一個儲存空間,儲存0,1,2的頻次 count = [0, 0, 0] n = len(nums) for i in range(n): count[nums[i]] += 1 index = 0 while count[0] > 0: nums[index] = 0 index += 1 count[0] -= 1 while count[1] > 0: nums[index] = 1 index += 1 count[1] -= 1 while count[2] > 0: nums[index] = 2 index += 1 count[2] -= 1
複雜度:
- 時間複雜度:O(n+n)
- 空間複雜度:O(k)
解法二:三路快排
由於是有重複的排序,並且只有三個數字,因此可以考慮三路快排序的partition思想
class Solution:
def sortColors(self, nums: List[int]) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
# [0,zero] -> 0
# [zero+1,i] -> 1
# [two,n-1] -> 2
n = len(nums)
zero = -1
two = n
i = 0
# 注意迴圈結束條件,到two即可,由於two後邊已經排列好
while i < two:
if nums[i] == 0:
zero += 1
nums[i],nums[zero] = nums[zero],nums[i]
i += 1
elif nums[i] == 1:
i += 1
else:
assert nums[i] == 2
two -= 1
nums[i],nums[two] = nums[two],nums[i]