You are given a?positive?integer array?nums.

Partition?nums?into two arrays,?nums1?and?nums2, such that:

• Each element of the array?nums?belongs to either the array?nums1?or the array?nums2.

• Both arrays are?non-empty.

• The value of the partition is?minimized.

The value of the partition is?|max(nums1) - min(nums2)|.

Here,?max(nums1)?denotes the maximum element of the array?nums1, and?min(nums2)?denotes the minimum element of the array?nums2.

Return?the integer denoting the value of such partition.

?

Example 1:

Input: nums = [1,3,2,4]Output: 1Explanation: We can partition the array nums into nums1 = [1,2] and nums2 = [3,4]. - The maximum element of the array nums1 is equal to 2. - The minimum element of the array nums2 is equal to 3. The value of the partition is |2 - 3| = 1. It can be proven that 1 is the minimum value out of all partitions.

Example 2:

Input: nums = [100,1,10]Output: 9Explanation: We can partition the array nums into nums1 = [10] and nums2 = [100,1]. - The maximum element of the array nums1 is equal to 10. - The minimum element of the array nums2 is equal to 1. The value of the partition is |10 - 1| = 9. It can be proven that 9 is the minimum value out of all partitions.

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Constraints:

• 2 <= nums.length <= 105

• 1 <= nums[i] <= 109

排序，計(jì)算每2個(gè)相鄰的差值，最小的即可返回；

Runtime:?22 ms, faster than?100.00%?of?Java?online submissions for?Find the Value of the Partition.

Memory Usage:?54.5 MB, less than?80.00%?of?Java?online submissions for?Find the Value of the Partition.