原題

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?

Reference Answer

思路分析

這道題本身不難，難點在要求一次遍歷完成排序。

這道題不允許使用排序庫函式。那麼最直觀的解法是：遍歷兩遍陣列，第一遍對0，1，2計數，第二遍對陣列進行賦值，這樣是可以ac的。但題目的要求是隻使用常數空間，而且只能遍歷一遍。那麼思路就比較巧妙了。設定兩個頭尾指標，頭指標p0指向的位置是0該放置的位置，尾指標p2指向的位置是2該放置的位置。i用來遍歷整個陣列，碰到0把它和p0指向的數交換，碰到2把它和p2指向的數交換，碰到1繼續向後遍歷。有點類似快速排序的分割陣列這一步。

class Solution:
    def sortColors
 
(self, nums):
        """
        :type nums: List[int]
        :rtype: void Do not return anything, modify nums in-place instead.
        """
#         if not nums:
#             return []
#         temp_list = [1 for i in range(len(nums))]
#         min_index = 0
#         max_index = len(nums) - 1
#         for x in nums:
#             if x == 0:
#                 temp_list[min_index] = 0
#                 min_index += 1
#             if x == 2:
#                 temp_list[max_index] = 2
#                 max_index -= 1
#         for index in range(len(nums)):
#             nums[index] = temp_list[index]
        
        if not nums:
            return []
        start_index = 0
        end_index = len(nums) - 1
        index = 0
        while index <= end_index:
            if nums[index] == 2:
                nums[end_index], nums[index] = nums[index], nums[end_index]
                end_index -= 1
            elif nums[index] == 0:
                nums[start_index], nums[index] = nums[index], nums[start_index]
                start_index += 1
                index += 1
            else:
                index += 1
        

Note:

• 本題在所含元素已知的前提下，完成一次遍歷排序的思想很值得借鑑，即設定頭指標p1，尾指標p2，i用來遍歷整個陣列，碰到0把它和p1指向的數交換，碰到2把它和p2指向的數交換，碰到1繼續向後遍歷。有點類似快速排序的分割陣列這一步。