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English Version

题目描述

给你一个整数数组 salary ,数组里每个数都是 唯一 的,其中 salary[i] 是第 i 个员工的工资。

请你返回去掉最低工资和最高工资以后,剩下员工工资的平均值。

 

示例 1:

输入:salary = [4000,3000,1000,2000]
输出:2500.00000
解释:最低工资和最高工资分别是 1000 和 4000 。
去掉最低工资和最高工资以后的平均工资是 (2000+3000)/2= 2500

示例 2:

输入:salary = [1000,2000,3000]
输出:2000.00000
解释:最低工资和最高工资分别是 1000 和 3000 。
去掉最低工资和最高工资以后的平均工资是 (2000)/1= 2000

示例 3:

输入:salary = [6000,5000,4000,3000,2000,1000]
输出:3500.00000

示例 4:

输入:salary = [8000,9000,2000,3000,6000,1000]
输出:4750.00000

 

提示:

  • 3 <= salary.length <= 100
  • 10^3 <= salary[i] <= 10^6
  • salary[i] 是唯一的。
  • 与真实值误差在 10^-5 以内的结果都将视为正确答案。

解法

方法一:模拟

按题意模拟即可。

遍历数组,求出最大值和最小值,并且累加和,然后求出去掉最大值和最小值后的平均值。

时间复杂度 $O(n)$。其中 $n$ 为数组 salary 的长度。

Python3

class Solution:
    def average(self, salary: List[int]) -> float:
        s = sum(salary) - min(salary) - max(salary)
        return s / (len(salary) - 2)

Java

class Solution {
    public double average(int[] salary) {
        int s = 0;
        int mi = 10000000, mx = 0;
        for (int v : salary) {
            mi = Math.min(mi, v);
            mx = Math.max(mx, v);
            s += v;
        }
        s -= (mi + mx);
        return s * 1.0 / (salary.length - 2);
    }
}

C++

class Solution {
public:
    double average(vector<int>& salary) {
        int s = 0;
        int mi = 1e7, mx = 0;
        for (int v : salary) {
            s += v;
            mi = min(mi, v);
            mx = max(mx, v);
        }
        s -= (mi + mx);
        return (double) s / (salary.size() - 2);
    }
};

Go

func average(salary []int) float64 {
	s := 0
	mi, mx := 10000000, 0
	for _, v := range salary {
		s += v
		mi = min(mi, v)
		mx = max(mx, v)
	}
	s -= (mi + mx)
	return float64(s) / float64(len(salary)-2)
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

TypeScript

function average(salary: number[]): number {
    let max = -Infinity;
    let min = Infinity;
    let sum = 0;
    for (const v of salary) {
        sum += v;
        max = Math.max(max, v);
        min = Math.min(min, v);
    }
    return (sum - max - min) / (salary.length - 2);
}

Rust

impl Solution {
    pub fn average(salary: Vec<i32>) -> f64 {
        let n = salary.len() as i32;
        let mut min = i32::MAX;
        let mut max = i32::MIN;
        let mut sum = 0;
        for &num in salary.iter() {
            min = min.min(num);
            max = max.max(num);
            sum += num;
        }
        f64::from(sum - min - max) / f64::from(n - 2)
    }
}

C

#define max(a,b) (((a) > (b)) ? (a) : (b))
#define min(a,b) (((a) < (b)) ? (a) : (b))

double average(int* salary, int salarySize) {
    int ma = INT_MIN;
    int mi = INT_MAX;
    int sum = 0;
    for (int i = 0 ; i < salarySize; i++) {
        sum += salary[i];
        ma = max(ma, salary[i]);
        mi = min(mi, salary[i]);
    }
    return (sum - mi - ma) * 1.0 / (salarySize - 2);
}

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