You are given a 0-indexed array nums
of n
integers, and an integer k
.
The k-radius average for a subarray of nums
centered at some index i
with the radius k
is the average of all elements in nums
between the indices i - k
and i + k
(inclusive). If there are less than k
elements before or after the index i
, then the k-radius average is -1
.
Build and return an array avgs
of length n
where avgs[i]
is the k-radius average for the subarray centered at index i
.
The average of x
elements is the sum of the x
elements divided by x
, using integer division. The integer division truncates toward zero, which means losing its fractional part.
- For example, the average of four elements
2
,3
,1
, and5
is(2 + 3 + 1 + 5) / 4 = 11 / 4 = 2.75
, which truncates to2
.
Example 1:
Input: nums = [7,4,3,9,1,8,5,2,6], k = 3 Output: [-1,-1,-1,5,4,4,-1,-1,-1] Explanation: - avg[0], avg[1], and avg[2] are -1 because there are less than k elements before each index. - The sum of the subarray centered at index 3 with radius 3 is: 7 + 4 + 3 + 9 + 1 + 8 + 5 = 37. Using integer division, avg[3] = 37 / 7 = 5. - For the subarray centered at index 4, avg[4] = (4 + 3 + 9 + 1 + 8 + 5 + 2) / 7 = 4. - For the subarray centered at index 5, avg[5] = (3 + 9 + 1 + 8 + 5 + 2 + 6) / 7 = 4. - avg[6], avg[7], and avg[8] are -1 because there are less than k elements after each index.
Example 2:
Input: nums = [100000], k = 0 Output: [100000] Explanation: - The sum of the subarray centered at index 0 with radius 0 is: 100000. avg[0] = 100000 / 1 = 100000.
Example 3:
Input: nums = [8], k = 100000 Output: [-1] Explanation: - avg[0] is -1 because there are less than k elements before and after index 0.
Constraints:
n == nums.length
1 <= n <= 105
0 <= nums[i], k <= 105
半径为 k 的子数组平均值。
给你一个下标从 0 开始的数组 nums ,数组中有 n 个整数,另给你一个整数 k 。
半径为 k 的子数组平均值 是指:nums 中一个以下标 i 为 中心 且 半径 为 k 的子数组中所有元素的平均值,即下标在 i - k 和 i + k 范围(含 i - k 和 i + k)内所有元素的平均值。如果在下标 i 前或后不足 k 个元素,那么 半径为 k 的子数组平均值 是 -1 。
构建并返回一个长度为 n 的数组 avgs ,其中 avgs[i] 是以下标 i 为中心的子数组的 半径为 k 的子数组平均值 。
x 个元素的 平均值 是 x 个元素相加之和除以 x ,此时使用截断式 整数除法 ,即需要去掉结果的小数部分。
例如,四个元素 2、3、1 和 5 的平均值是 (2 + 3 + 1 + 5) / 4 = 11 / 4 = 2.75,截断后得到 2 。
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/k-radius-subarray-averages
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。
思路是前缀和。求一段子数组的和的做法是前缀和,这应该成为刷题的本能反应。
我们首先用一个额外数组记录整个数组的前缀和,然后再次遍历数组,根据条件计算以每一个下标 i 为中心,半径为 k 的子数组的和。
时间O(n)
空间O(n)
Java实现
1 class Solution { 2 public int[] getAverages(int[] nums, int k) { 3 int n = nums.length; 4 int[] res = new int[n]; 5 Arrays.fill(res, -1); 6 // 要用long型,不然超出范围 7 long[] preSum = new long[n + 1]; 8 for (int i = 0; i < n; i++) { 9 preSum[i + 1] = preSum[i] + nums[i]; 10 } 11 12 for (int i = 0; i < nums.length; i++) { 13 if (i - k >= 0 && i + k < n) { 14 res[i] = (int)((preSum[i + k + 1] - preSum[i - k]) / (2 * k + 1)); 15 } 16 } 17 return res; 18 } 19 }
标签:index,elements,nums,int,Radius,数组,Subarray,avg,Averages From: https://www.cnblogs.com/cnoodle/p/17493688.html