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1619. 删除某些元素后的数组均值

English Version

题目描述

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。

 

示例 1：

输入：arr = [1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3]
输出：2.00000
解释：删除数组中最大和最小的元素后，所有元素都等于 2，所以平均值为 2 。

示例 2：

输入：arr = [6,2,7,5,1,2,0,3,10,2,5,0,5,5,0,8,7,6,8,0]
输出：4.00000

示例 3：

输入：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]
输出：4.77778

示例 4：

输入：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]
输出：5.27778

示例 5：

输入：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]
输出：5.29167

 

提示：

  • 20 <= arr.length <= 1000
  • arr.length 是 20 的 倍数 
  • 0 <= arr[i] <= 105

解法

方法一：模拟

直接模拟。

先对数组 arr 排序，然后截取中间的 90% 个元素，求平均值。

时间复杂度 $O(n\log n)$。其中 $n$ 为数组 arr 的长度。

Python3

class Solution:
    def trimMean(self, arr: List[int]) -> float:
        n = len(arr)
        start, end = int(n * 0.05), int(n * 0.95)
        arr.sort()
        t = arr[start:end]
        return round(sum(t) / len(t), 5)

Java

class Solution {
    public double trimMean(int[] arr) {
        Arrays.sort(arr);
        int n = arr.length;
        double s = 0;
        for (int start = (int) (n * 0.05), i = start; i < n - start; ++i) {
            s += arr[i];
        }
        return s / (n * 0.9);
    }
}

TypeScript

function trimMean(arr: number[]): number {
    arr.sort((a, b) => a - b);
    let n = arr.length,
        rmLen = n * 0.05;
    let sum = 0;
    for (let i = rmLen; i < n - rmLen; i++) {
        sum += arr[i];
    }
    return sum / (n * 0.9);
}

C++

class Solution {
public:
    double trimMean(vector<int>& arr) {
        sort(arr.begin(), arr.end());
        int n = arr.size();
        double s = 0;
        for (int start = (int)(n * 0.05), i = start; i < n - start; ++i)
            s += arr[i];
        return s / (n * 0.9);
    }
};

Go

func trimMean(arr []int) float64 {
	sort.Ints(arr)
	n := len(arr)
	sum := 0.0
	for i := n / 20; i < n-n/20; i++ {
		sum += float64(arr[i])
	}
	return sum / (float64(n) * 0.9)
}

Rust

impl Solution {
    pub fn trim_mean(mut arr: Vec<i32>) -> f64 {
        arr.sort();
        let n = arr.len();
        let count = (n as f64 * 0.05).floor() as usize;
        let mut sum = 0;
        for i in count..n - count {
            sum += arr[i];
        }
        sum as f64 / (n as f64 * 0.9)
    }
}

...