You are given a 0-indexed integer array nums, where nums[i] represents the score of the ith student. You are also given an integer k.
Pick the scores of any k students from the array so that the difference between the highest and the lowest of the k scores is minimized.
Return the minimum possible difference.
Example 1:
Input: nums = [90], k = 1
Output: 0
Explanation: There is one way to pick score(s) of one student:
- [90]. The difference between the highest and lowest score is 90 - 90 = 0.
The minimum possible difference is 0.
Example 2:
Input: nums = [9,4,1,7], k = 2 Output: 2 Explanation: There are six ways to pick score(s) of two students: - [9,4,1,7]. The difference between the highest and lowest score is 9 - 4 = 5. - [9,4,1,7]. The difference between the highest and lowest score is 9 - 1 = 8. - [9,4,1,7]. The difference between the highest and lowest score is 9 - 7 = 2. - [9,4,1,7]. The difference between the highest and lowest score is 4 - 1 = 3. - [9,4,1,7]. The difference between the highest and lowest score is 7 - 4 = 3. - [9,4,1,7]. The difference between the highest and lowest score is 7 - 1 = 6. The minimum possible difference is 2.
Constraints:
1 <= k <= nums.length <= 10000 <= nums[i] <= 105
学生分数的最小差值。
给你一个 下标从 0 开始 的整数数组 nums ，其中 nums[i] 表示第 i 名学生的分数。另给你一个整数 k 。
从数组中选出任意 k 名学生的分数，使这 k 个分数间 最高分 和 最低分 的 差值 达到 最小化 。
返回可能的 最小差值 。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/minimum-difference-between-highest-and-lowest-of-k-scores
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
这道题的思路是排序 + 滑动窗口。这里有一个 corner case 需要排除，就是当 k = 1 的时候，因为最大值和最小值是同一个数字，所以结果是 0。对于一般的情况，因为排完序之后，左指针指向的是最小的元素，右指针指向的是最大的元素，所以只要卡好左右指针的距离一直等于 k，我们就可以扫描一遍数组从而得到所有 k 个数字中间那个最小的差值。
时间O(nlogn)
空间O(1)
Java实现
1 class Solution { 2 public int minimumDifference(int[] nums, int k) { 3 // corner case 4 if (k == 1) { 5 return 0; 6 } 7 8 // normal case 9 Arrays.sort(nums); 10 int res = Integer.MAX_VALUE; 11 for (int i = k - 1; i < nums.length; i++) { 12 res = Math.min(res, nums[i] - nums[i - k + 1]); 13 } 14 return res; 15 } 16 }