LeetCode1491. Average value of wages after removing the minimum wage and the maximum wage

【 Simple 】 Give you an array of integers salary , Every number in the array is only Of , among salary[i] It's No i The salary of an employee .

Please go back and remove the minimum wage and the maximum wage , The average wage of the remaining employees .

Example 1：

 Input ：salary = [4000,3000,1000,2000]
 Output ：2500.00000
 explain ： The minimum wage and the maximum wage are respectively  1000  and  4000 .
 The average wage after removing the minimum wage and the maximum wage is  (2000+3000)/2= 2500

Example 2：

 Input ：salary = [1000,2000,3000]
 Output ：2000.00000
 explain ： The minimum wage and the maximum wage are respectively  1000  and  3000 .
 The average wage after removing the minimum wage and the maximum wage is  (2000)/1= 2000

Example 3：

 Input ：salary = [6000,5000,4000,3000,2000,1000]
 Output ：3500.00000

Example 4：

 Input ：salary = [8000,9000,2000,3000,6000,1000]
 Output ：4750.00000

Tips ：

3 <= salary.length <= 100
10^3 <= salary[i] <= 10^6
salary[i] Is the only one. .
The error with the true value is within 10^-5 All the results within will be regarded as the correct answer .

Solution 1 ：

double m1(int[] salary) {
    
        Arrays.sort(salary);
        int sum = 0;
        for (int i = 1; i < salary.length-1; i++) {
    
            sum += salary[i];
        }
        return (double) sum / (salary.length-2);
    }

Solution 2 ：

double m2(int[] salary) {
    
        int max = salary[0];
        int min = salary[0];
        int sum = 0;
        for (int i = 0; i < salary.length; i++) {
    
            sum += salary[i];
            if (salary[i] > max)
                max = salary[i];
            else if (salary[i] < min)
                min = salary[i];
        }
        return (double) (sum-max-min) / (salary.length-2);
    }