LeetCode 75. Sort Colors (颜色排序)

原题

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

这道题本身不难，难点在要求一次遍历完成排序。

这道题不允许使用排序库函数。那么最直观的解法是：遍历两遍数组，第一遍对0，1，2计数，第二遍对数组进行赋值，这样是可以ac的。但题目的要求是只使用常数空间，而且只能遍历一遍。那么思路就比较巧妙了。设置两个头尾指针，头指针p0指向的位置是0该放置的位置，尾指针p2指向的位置是2该放置的位置。i用来遍历整个数组，碰到0把它和p0指向的数交换，碰到2把它和p2指向的数交换，碰到1继续向后遍历。有点类似快速排序的分割数组这一步。

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:

• 本题在所含元素已知的前提下，完成一次遍历排序的思想很值得借鉴，即设置头指针p1，尾指针p2，i用来遍历整个数组，碰到0把它和p1指向的数交换，碰到2把它和p2指向的数交换，碰到1继续向后遍历。有点类似快速排序的分割数组这一步。