mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-07 01:44:56 +08:00
74 lines
1.9 KiB
Markdown
74 lines
1.9 KiB
Markdown
# [1619. Mean of Array After Removing Some Elements](https://leetcode.com/problems/mean-of-array-after-removing-some-elements/)
|
||
|
||
## 题目
|
||
|
||
Given an integer array `arr`, return *the mean of the remaining integers after removing the smallest `5%` and the largest `5%` of the elements.*
|
||
|
||
Answers within `10-5` of the **actual answer** will be considered accepted.
|
||
|
||
**Example 1:**
|
||
|
||
```
|
||
Input: arr = [1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3]
|
||
Output: 2.00000
|
||
Explanation: After erasing the minimum and the maximum values of this array, all elements are equal to 2, so the mean is 2.
|
||
```
|
||
|
||
**Example 2:**
|
||
|
||
```
|
||
Input: arr = [6,2,7,5,1,2,0,3,10,2,5,0,5,5,0,8,7,6,8,0]
|
||
Output: 4.00000
|
||
```
|
||
|
||
**Example 3:**
|
||
|
||
```
|
||
Input: arr = [6,0,7,0,7,5,7,8,3,4,0,7,8,1,6,8,1,1,2,4,8,1,9,5,4,3,8,5,10,8,6,6,1,0,6,10,8,2,3,4]
|
||
Output: 4.77778
|
||
```
|
||
|
||
**Example 4:**
|
||
|
||
```
|
||
Input: arr = [9,7,8,7,7,8,4,4,6,8,8,7,6,8,8,9,2,6,0,0,1,10,8,6,3,3,5,1,10,9,0,7,10,0,10,4,1,10,6,9,3,6,0,0,2,7,0,6,7,2,9,7,7,3,0,1,6,1,10,3]
|
||
Output: 5.27778
|
||
```
|
||
|
||
**Example 5:**
|
||
|
||
```
|
||
Input: arr = [4,8,4,10,0,7,1,3,7,8,8,3,4,1,6,2,1,1,8,0,9,8,0,3,9,10,3,10,1,10,7,3,2,1,4,9,10,7,6,4,0,8,5,1,2,1,6,2,5,0,7,10,9,10,3,7,10,5,8,5,7,6,7,6,10,9,5,10,5,5,7,2,10,7,7,8,2,0,1,1]
|
||
Output: 5.29167
|
||
```
|
||
|
||
**Constraints:**
|
||
|
||
- `20 <= arr.length <= 1000`
|
||
- `arr.length` **is a multiple** of `20`.
|
||
- `0 <= arr[i] <= 10^5`
|
||
|
||
## 题目大意
|
||
|
||
给你一个整数数组 `arr` ，请你删除最小 `5%` 的数字和最大 `5%` 的数字后，剩余数字的平均值。与 **标准答案** 误差在 `10-5` 的结果都被视为正确结果。
|
||
|
||
## 解题思路
|
||
|
||
- 简单题。先将数组排序，然后累加中间 90% 的元素，最后计算平均值。
|
||
|
||
## 代码
|
||
|
||
```go
|
||
package leetcode
|
||
|
||
import "sort"
|
||
|
||
func trimMean(arr []int) float64 {
|
||
sort.Ints(arr)
|
||
n, sum := len(arr), 0
|
||
for i := n / 20; i < n-(n/20); i++ {
|
||
sum += arr[i]
|
||
}
|
||
return float64(sum) / float64((n - (n / 10)))
|
||
}
|
||
``` |