[leetcode -- the first day of introduction to programming ability] basic data type [statistics of odd numbers within the range / average wage after removing the minimum wage and maximum wage)

Catalog

subject ： Count the odd number in the interval

Error model ：

analysis ：

  Problem solving ： Count the odd number in the interval

subject ： The average wage after the minimum wage and the maximum wage are removed

  Error model ：

For Traverse

Arrays class

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subject ： Count the odd number in the interval

Here are two nonnegative integers  low and  high . Please return  low and  high  Between （ Including both ） An odd number .

Example 1：

Input ：low = 3, high = 7
Output ：3
explain ：3 To 7 The odd number in between is [3,5,7] .
Example 2：

Input ：low = 8, high = 10
Output ：1
explain ：8 To 10 The odd number in between is [9] .

Tips ：

0 <= low <= high <= 10^9.

Error model ：

class Solution {
    public int countOdds(int low, int high) {
    int sum=0;// Cumulative sum 
// from low To hign Traverse in turn 
    for(int i=low;i<=high;i++){
// If it is judged to be odd, it will be added by itself 
        if(i%2!=0){
 sum++;
        } 
    }
    return sum;
    }
}

The first thought of direct loop traversal is too lazy to use your brain , After clicking the test, I found Beyond the time limit . Remember, this should be an algorithm problem .

analysis ：

I think so , Every number is an odd number 、 even numbers 、 Odd number 、 even numbers 、 This cycle . There are three possibilities low and height If both are odd numbers, use (high-low)/2+1, If both are even numbers (high-low)/2, If both are an odd number, an even number can be used (high-low+1)/2.

Both odd numbers ：

Both are even numbers

  Odd and even

  Problem solving ： Count the odd number in the interval

class Solution {
    public int countOdds(int low, int high) {
    if(low%2!=0&&high%2!=0){
        return (high-low)/2+1;
    }else if(low%2==0&&high%2==0){
     return (high-low)/2;
    }else if(low%2!=0&&high%2==0||low%2==0&&high%2!=0){
     return (high-low+1)/2;
    }
return 0;
}
}

subject ： The average wage after the minimum wage and the maximum wage are removed

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 .

  Error model ：

No,

analysis : We use Arrays class Of sort The sorting method is OK for Loop traversal can also be done without problems

For Traverse

class Solution {
    public double average(int[] salary) {
        double min=salary[0];// Storage minimum 
        double max=salary[0];// Storage maximum 
        double sum=0;// Store accumulation and 
     for(int i=0;i<salary.length;i++){
         if(salary[i]>max){
             max=salary[i];
         }
         if(salary[i]<min){
             min=salary[i];
         }
// Accumulate and store 
         sum+=salary[i];
     }
// Subtract the maximum and minimum values 
     sum=sum-max-min;
     return sum/(salary.length-2);
    }
}

Arrays class

class Solution {
    public double average(int[] salary) {
// Sort automatically 
     Arrays.sort(salary);
// Store the accumulated value 
     double a=0;
// The maximum and minimum values are set to 0
     salary[0]=0;
     salary[salary.length-1]=0;
     for(int i=0;i<salary.length;i++){
     a+=salary[i];
     }
     return a/(salary.length-2);
    }
}

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