题目：
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
算法解析：
首先对数组里面进行排序，那么在不改变数字的情况下，最大的差值为nums[0] - nums[nums.size()-1]
那么，在改变三个值的情况下，改变的是前缀和后缀，有以下几种情况：
前缀：0 后缀：3 nums[n - 1 - 3] - nums[0]
前缀：1 后缀：2 nums[n - 1 - 2] - nums[1]
前缀：2 后缀：1 nums[n - 1 - 1] - nums[2]
前缀：3 后缀：0 nums[n - 1] - nums[3]
代码如下：
class Solution {
public:int minDifference(vector<int>& nums) {int n = nums.size();if(n <= 3) return 0;sort(nums.begin(),nums.end());int res = INT_MAX;for(int i = 0, j = 3; i <= 3; i++,j--){res = min(res,nums[n - 1 - j] - nums[i]);}return res;}
};