Maximum Gap Leetcode #164 题解[Python]

题目来源

https://leetcode.com/problems/maximum-gap/description/

题意分析

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Return 0 if the array contains less than 2 elements.

Example 1:

Input: [3,6,9,1]
Output: 3
Explanation: The sorted form of the array is [1,3,6,9], either
(3,6) or (6,9) has the maximum difference 3.

Example 2:

Input: [10]
Output: 0
Explanation: The array contains less than 2 elements, therefore return 0.

Note:

You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
Try to solve it in linear time/space.

给出一个乱序数组, 求出排序后连续的两个元素间最大差值, 这个差值即是所谓的MaxGap.

元素少于2返回0
尝试在线性空间\时间内解决.

解题思路

这题坦白来讲其实相当的有难度. 博主想了挺久也没有想出可行的线性解.

后来看了讨论区, 原来这道题用桶排序变形一下就好了.

首先, 对一个数组来说, 肯定会有这个公式成立: $MaxGap>=\lceil(max-min)/(n-2)\rceil$ , 即MaxGap会大于等于排好序的数组的平均间隔, 我们不妨称这个平均间隔为average_gap.

那么我们依据这个average_gap将乱序数组划分为n-2个桶, MaxGap便只可能在相邻的两个桶, 前一个桶的最大值与后一个桶的最小值的差值中产生.

代码实现

import sys
import math


class Solution:
    def maximumGap(self, nums):
        """
        : type nums : List[int]
        : rtype : int
        """
        if len(nums) < 2:
            return 0
        if len(nums)==2:
            return int(math.fabs(nums[0]-nums[1]))
        Min = sys.maxsize
        Max = -sys.maxsize
        for num in nums:
            if num < Min:
                Min = num
            if num > Max:
                Max = num


        # Spilt the array into buckets
        average_gap = int(math.ceil((Max - Min) / (len(nums) - 2)))
        buckets = [[] for i in range(len(nums) - 2)]
        for num in nums:
            if num == Min or num == Max:
                continue
            else:
                idx = int(math.floor((num - Min) / average_gap))
                buckets[idx].append(num)

        # Get the bucket min and bucket max
        pre = Min
        ret = -sys.maxsize
        for bucket in buckets:
            if not bucket:
                pass
            else:
                ret = max(min(bucket) - pre, ret)
                pre = max(bucket)
        return max(ret,Max-pre)

代码表现

代码表现差强人意. 后来反思了一下还可以做点小优化, 那就是用空间换时间, 事先保存好每个桶的最大值和最小值, 这样每次把新的元素加入桶中时都维护好最大值和最小值.

汇总求MaxGap时便可以直接拿来用而不用再用一遍max和min方法.

164_ac