【LeetCode with Python】 75. Sort Colors

题目

原题页面:https://oj.leetcode.com/problems/sort-colors/
本文地址:http://leetcode.xnerv.wang/sort-colors/
题目类型:Array, Two Pointers, Sort
难度评价:Medium
类似题目:(M) Sort List, (M) Wiggle Sort

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.

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 an one-pass algorithm using only constant space?


分析

定义两个两头的下标left和right,i从左至右遍历数组,发现0就和left交换(同时left+=1),发现2就和right交换(同时right-=1)。这样就能确保下标 < left的元素都是0,而下标 > right的元素都是2。
还可以进行进一步的优化,left指针每次都移动到一个不是0的位置,right指针每次都移动到一个不是2的位置,这样也许可以减少一些交换次数。


代码

class Solution:
    # @param A a list of integers
    # @return nothing, sort in place
    def sortColors(self, A):
        len_A = len(A)
        if 1 == len(A):
            return
        left = 0
        right = len_A - 1
        i = 0
        while i <= right:
            if 0 == A[i]:
                if left == i:
                    i += 1
                else:
                    A[left], A[i] = A[i], A[left]
                left += 1
            elif 1 == A[i]:
                i += 1
            else:
                if right == i:
                    i += 1
                else:
                    A[right], A[i] = A[i], A[right]
                right -= 1

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