[leetcode:python]75.Sort Colors

最新推荐文章于 2019-06-17 21:39:46 发布

原创最新推荐文章于 2019-06-17 21:39:46 发布 · 966 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#python #leetcode

算法专栏收录该内容

150 篇文章

订阅专栏

本文介绍了一种不使用库函数的三色排序算法，通过将数组中的0、1、2分别代表红、白、蓝三种颜色并按顺序排列，提供两种实现方法：一种是通过将0放到前面，2放到后面实现；另一种是利用Lomuto分区算法，通过一次遍历完成排序。

题目：
Given an array with n objects colored red, white or blue, sort them 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.
题意：
用0，1，2表示不同的三种颜色，给定一个颜色序列，写程序使得相同的颜色处于相邻的位置。
不能使用库的函数。

方法一：性能42ms
把0都放前面，2都放后面

class Solution(object):
    def sortColors(self, nums):
        """
        :type nums: List[int]
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        if nums== []:
            return 
        p0 = 0
        p2 = len(nums) - 1
        i = 0
        while i <= p2:
            if nums[i] == 2:
                nums[i], nums[p2] = nums[p2], nums[i]
                p2 -= 1
            elif nums[i] == 0:
                nums[i], nums[p0] = nums[p0], nums[i]
                p0 +=1
                i += 1
            else:
                i += 1

方法二：性能35ms

class Solution(object):
    def sortColors(self, nums):
        """
        :type nums: List[int]
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        # Using Lomuto's partition algorithm
        # store value for each iteration
        # In-place one pass solution O(n) O(1)
        # No swap required
        # [0, i) are 0's
        # [i, j) are 1's
        # [j, k) are 2's
        i, j = 0, 0
        for k in range(len(nums)):
            cur = nums[k]   # store value
            nums[k] = 2 # make color blue
            if cur < 2: # if red or white
                nums[j] = 1 # make white
                j += 1
            if cur == 0:
                nums[i] = 0
                i += 1