Java : 배열에서 가장 높은 값을 찾는 것

Question

어떤 이유로이 코드는 하나만 인쇄하려고 할 때 배열에서 가장 높은 값에 대해 세 값을 인쇄하는 것입니다 (11.3). 누군가 왜이 일을하는지 설명해 주시겠습니까?

감사.

import java.util.Scanner;

public class Slide24
{
    public static void main (String [] args)
    {
        Scanner in = new Scanner(System.in);

        double[] decMax = {-2.8, -8.8, 2.3, 7.9, 4.1, -1.4, 11.3, 10.4,
            8.9, 8.1, 5.8, 5.9, 7.8, 4.9, 5.7, -0.9, -0.4, 7.3, 8.3, 6.5, 9.2,
            3.5, 3, 1.1, 6.5, 5.1, -1.2, -5.1, 2, 5.2, 2.1};

        double total = 0, avgMax = 0;

        for (int counter = 0; counter < decMax.length; counter++)
        {
         total += decMax[counter];
        }

        avgMax = total / decMax.length;

        System.out.printf("%s %2.2f\n", "The average maximum temperature for December was: ", avgMax);

        //finds the highest value
        double max = decMax[0];

        for (int counter = 1; counter < decMax.length; counter++)
        {
         if (decMax[counter] > max)
         {
          max = decMax[counter];
          System.out.println("The highest maximum for the December is: " + max);
         }

        }        
    }
}

Answer 1

It's printing out a number every time it finds one that is higher than the current max (which happens to occur three times in your case.) Move the print outside of the for loop and you should be good.

for (int counter = 1; counter < decMax.length; counter++)
{
     if (decMax[counter] > max)
     {
      max = decMax[counter];
     }
}

System.out.println("The highest maximum for the December is: " + max);

Answer 2

To find the highest (max) or lowest (min) value from an array, this could give you the right direction. Here is an example code for getting the highest value from a primitive array.

Method 1:

public int maxValue(int array[]){
  List<Integer> list = new ArrayList<Integer>();
  for (int i = 0; i < array.length; i++) {
    list.add(array[i]);
  }
 return Collections.max(list);

}

To get the lowest value, you can use

Collections.min(list)

Method 2:

public int maxValue(int array[]){
  int max = Arrays.stream(array).max().getAsInt();
  return max;
}

Now the following line should work.

System.out.println("The highest maximum for the December is: " + maxValue(decMax)); 

Answer 3

You need to print out the max after you've scanned all of them:

for (int counter = 1; counter < decMax.length; counter++)
{
    if (decMax[counter] > max)
    {
        max = decMax[counter];
        // not here: System.out.println("The highest maximum for the December is: " + max);
    }
}  
System.out.println("The highest maximum for the December is: " + max);

Answer 4

Same as suggested by others, just mentioning the cleaner way of doing it:

int max = decMax[0];
for(int i=1;i<decMax.length;i++)
    max = Math.max(decMax[i],max);
System.out.println("The Maximum value is : " + max);

Answer 5

If you are looking for the quickest and simplest way to perform various actions in regards to arrays, the use of the Collections class is extremely helpful (documentation available from https://docs.oracle.com/javase/7/docs/api/java/util/Collections.html), actions ranges from finding the maximum, minimum, sorting, reverse order, etc.

A simple way to find the maximum value from the array with the use of Collections:

Double[] decMax = {-2.8, -8.8, 2.3, 7.9, 4.1, -1.4, 11.3, 10.4, 8.9, 8.1, 5.8, 5.9, 7.8, 4.9, 5.7, -0.9, -0.4, 7.3, 8.3, 6.5, 9.2, 3.5, 3.0, 1.1, 6.5, 5.1, -1.2, -5.1, 2.0, 5.2, 2.1};
List<Double> a = new ArrayList<Double>(Arrays.asList(decMax));
System.out.println("The highest maximum for the December is: " + Collections.max(a));

If you are interested in finding the minimum value, similar to finding maximum:

System.out.println(Collections.min(a));

The simplest line to sort the list:

Collections.sort(a);

Or alternatively the use of the Arrays class to sort an array:

Arrays.sort(decMax);

However the Arrays class does not have a method that refers to the maximum value directly, sorting it and referring to the last index is the maximum value, however keep in mind sorting by the above 2 methods has a complexity of O(n log n).

Answer 6

A shorter solution to have the max value of array:

double max = Arrays.stream(decMax).max(Double::compareTo).get();

Answer 7

You have your print() statement in the for() loop, It should be after so that it only prints once. the way it currently is, every time the max changes it prints a max.

Answer 8

You are just Compare zeroth element with rest of the elements so it will print the value last largest value which it will contain, in you case, same problem is happening. To compare each and every elements we have to swap the values like:

double max = decMax[0];
    for (int counter = 1; counter < decMax.length; counter++){
        if(max<decMax[i]){
            max=decMax[i]; //swapping
            decMax[i]=decMax[0];
        }
    }
    System.out.println("The max value is "+ max);

Hope this will help you

Answer 9

You can use a function that accepts a array and finds the max value in it. i made it generic so it could also accept other data types

public static <T extends Comparable<T>> T findMax(T[] array){       
    T max = array[0];
    for(T data: array){
        if(data.compareTo(max)>0)
            max =data;                
    }
    return max;
}

Answer 10

You can write like this.

import java.util.Scanner;
class   BigNoArray{

    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter how many array element");
        int n=sc.nextInt();
        int[] ar= new int[n];
        System.out.println("enter "+n+" values");
        for(int i=0;i<ar.length;i++){
            ar[i]=sc.nextInt();
        }
        int fbig=ar[0];
        int sbig=ar[1];
        int tbig=ar[3];
            for(int i=1;i<ar.length;i++){
                if(fbig<ar[i]){
                    sbig=fbig;
                    fbig=ar[i];
                }
                else if(sbig<ar[i]&&ar[i]!=fbig){
                    sbig=ar[i];
                }
                else if(tbig<ar[i]&&ar[i]!=fbig){
                    tbig=ar[i];
                }
            }
        System.out.println("first big number is "+fbig);
        System.out.println("second big number is "+sbig);
        System.out.println("third big number is "+tbig);
    }
}

Answer 11

void FindMax()
{
    int lessonNum;

    System.out.print("Enter your lesson numbers : ");
    lessonNum = input.nextInt();
    int[] numbers = new int[lessonNum];
    for (int i = 0; i < numbers.length; i++)
    {
        System.out.print("Please enter " + (i + 1) + " number : ");
        numbers[i] = input.nextInt();
    }
    double max = numbers[0];
    for (int i = 1; i < numbers.length; i++)
    {
        if (numbers[i] > max)
        {
            max = numbers[i];
        }
    }
    System.out.println("Maximum number is : " + max);
}

Answer 12

Easiest way which I've found, supports all android versions

Arrays.sort(series1Numbers);

int maxSeries = Integer.parseInt(String.valueOf(series1Numbers[series1Numbers.length-1]));

Answer 13

An easy way without using collections

public void findHighestNoInArray() {
        int[] a = {1,2,6,8,9};
        int large = a[0];
            for(int num : a) {
                if(large < num) {
                    large = num;
                }
            }
            System.out.println("Large number is "+large+"");
        }

Answer 14

If you don't want to use any java predefined libraries then below is the

simplest way

   public class Test {

    public static void main(String[] args) {
        double[] decMax = {-2.8, -8.8, 2.3, 7.9, 4.1, -1.4, 11.3, 10.4,
            8.9, 8.1, 5.8, 5.9, 7.8, 4.9, 5.7, -0.9, -0.4, 7.3, 8.3, 6.5, 9.2,
            3.5, 3, 1.1, 6.5, 5.1, -1.2, -5.1, 2, 5.2, 2.1};

        double maxx = decMax[0];

        for (int i = 0; i < decMax.length; i++) {
            if (maxx < decMax[i]) {
                maxx = decMax[i];
            }
        }
        System.out.println(maxx);

    }
}

Answer 15

import java.util.*;
class main9 //Find the smallest and 2lagest and  ascending and descending order of  elements in array//
{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        System.out.println("Enter the array range");
        int no=sc.nextInt();
        System.out.println("Enter the array element");
        int a[]=new int[no];
        int i;
        for(i=0;i<no;i++)
        {
            a[i]=sc.nextInt();
        }
        Arrays.sort(a);
        int s=a[0];
        int l=a[a.length-1];
        int m=a[a.length-2];
        System.out.println("Smallest no is="+s);
        System.out.println("lagest 2 numbers are=");
        System.out.println(l);
        System.out.println(m);
        System.out.println("Array in ascending:");
        for(i=0;i<no;i++)
        {
            System.out.println(a[i]);
        }
        System.out.println("Array in descending:");
        for(i=a.length-1;i>=0;i--)
        {
            System.out.println(a[i]);
        }
    }
}