動機

  • 忘了bucket sort
  • 在python用[0]*n通常下場都不好，要改用dict
  • sort還是比較厲害

Problem

Given an integer array nums, return the maximum difference between two successive elements in its sorted form. If the array contains less than two elements, return 0.

You must write an algorithm that runs in linear time and uses linear extra space.

 

Example 1:

Input: nums = [3,6,9,1]Output: 3Explanation: 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: nums = [10]Output: 0Explanation: The array contains less than 2 elements, therefore return 0.

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 109

Sol

bucket sort就是把list，分堆，放到bucket中

所以這裡要根據數字的範圍，分到array的孔去，就是下面的B_len 但要注意可能是0，像[1,1,1]

class Solution:
    def maximumGap(self, nums):
        if len(nums) <= 2:
            return max(nums)-min(nums)
        mi = min(nums)
        nums = [n-mi for n in nums]
        val_len = max(nums)
        B_len = val_len//len(nums)
        B_len = B_len if B_len > 0 else 1
        Bmax = defaultdict(lambda : float('-inf'))
        Bmin = defaultdict(lambda : float('inf'))
        for n in nums:
            i = n//B_len
            Bmax[i] = max(Bmax[i], n)
            Bmin[i] = min(Bmin[i], n)
        
        ret,pre = 0,0
        for i in range(1,len(nums)+1):
            if i in Bmin:
                ret = max(ret, Bmin[i]-Bmax[pre])
                pre = i
        return ret