【leetcode】1509. Minimum Difference Between Largest and Smallest Value in Three Moves

题目如下：

Given an array nums, you are allowed to choose one element of nums and change it by any value in one move.

Return the minimum difference between the largest and smallest value of nums after perfoming at most 3 moves. 

Example 1:

Input: nums = [5,3,2,4]
Output: 0
Explanation: Change the array [5,3,2,4] to [2,2,2,2].
The difference between the maximum and minimum is 2-2 = 0.

Example 2:

Input: nums = [1,5,0,10,14]
Output: 1
Explanation: Change the array [1,5,0,10,14] to [1,1,0,1,1]. 
The difference between the maximum and minimum is 1-0 = 1.

Example 3:

Input: nums = [6,6,0,1,1,4,6]
Output: 2

Example 4:

Input: nums = [1,5,6,14,15]
Output: 1

Constraints:

• 1 <= nums.length <= 10^5
• -10^9 <= nums[i] <= 10^9

解题思路：本题有点意思。要使得最大值和最小值的差最小，那么就要尽量使最小值变大，最大值变小。所以可以分成如下几种情况，一是修改最小的三个值，二是修改最小的两个值以及一个最大值，三是修改最小值以及两个最大值，四是修改最大的三个值，最后比较这四种情况的结果即可。

代码如下：

class Solution(object):
    def minDifference(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums.sort()
        if len(nums) <= 3:
            return 0
        return min( abs(nums[-1] - nums[3]),abs(nums[-2] - nums[2]),abs(nums[-3] - nums[1]),abs(nums[-4] - nums[0]) )