题目 #

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: 1
Explanation: 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: 9
Explanation: 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.

Constraints:

• $2 <= nums.length <= 10^5$
• $1 <= nums[i] <= 10^9$

思路1 分析+排序 #

分析 #

• 看似分两个数组，规定了nums1取max，nums2取min，那么找到两个相邻最近的数字a >= b
• 比b小的都放到nums1中，最大值就是b
• 比a大的都放到nums2中，最小值就是a
• 结果就是a - b
• 排序找最小间隔返回即可

代码 #

 1func findValueOfPartition(nums []int) int {
 2	sort.Ints(nums)
 3	res := math.MaxInt
 4	for i := range nums[1:] {
 5		diff := nums[i+1] - nums[i]
 6		if diff < res {
 7			res = diff
 8		}
 9	}
10	return res
11}