Question

我无法解决此家庭作业问题。问题要我创建一个程序来读取用户输入的数字并获取这些数字的最小值和最大值。

基本上，输出应该如下：

输入电话号码：10

输入10个以空格分隔的数字，然后按Enter：1 2 3 1 2 3 4 5 6 3

最小值为1且有2次出现

最大为6，并且出现1次

我能够创建方法来获取最小值和最大值。我不知道如何用我所拥有的最小和最大次数。我也不知道如何让扫描仪读取同一行上的整数输入。

import java.util.Scanner;

public class homework2
{
  public int min(int[] array)
  {
    int min = array[0];
    for (int i = 0; i < array.length; i++)
    {
      if (array[i] < min)
      {
        min = array[i];
      }
    }
    return min;
  }

  public int max(int[] array)
  {
    int max = 0;
    for (int i = 0; i < array.length; i++)
    {
      if (array[i] > max)
      {
        max = array[i];
      }
    }
    return max;
  }

  public static void main(String[] args)
  {
    Scanner s = new Scanner(System.in);
    System.out.println("Enter the length of the array:");
    int length = s.nextInt();
    int[] myArray = new int[length];
    System.out.println("Enter the elements of the array:");

    for (int i = 0; i < length; i++)
    {
      myArray[i] = s.nextInt();
    }
  }
}

Answer 1

一种方法是存储使用HashMap看到的每个数字的计数。数字本身可以是键，值可以是您要递增的计数。您还可以使用Math lib从一组键（数字）中输出最小值和最大值

https://www.geeksforgeeks.org/java-math-min-method-examples/

https://www.geeksforgeeks.org/count-occurrences-elements-list-java/

祝你好运！

Answer 2

一旦知道最大数量，只需计算数组中出现的次数。

但是，正如@Babyburger指出的那样，您不需要一个数组即可计算最大值或出现的次数。

Answer 3

这是一个简单的程序，可满足您的需求：

public static void main(String[] args) {
    int[] array = {1, 2, 3, 1, 2, 3, 4, 5, 6, 3}; // get your actual array
    int first = array[0];

    // initial values
    int min = first;
    int minOccurs = 1;
    int max = first;
    int maxOccurs = 1;

    for(int i = 1; i < array.length; i++) {
        int current = array[i];
        if(current == min) {
            minOccurs++;
        } else if (current < min) {
            min = current;
            minOccurs = 1;
        }

        if(current == max) {
            maxOccurs++;
        } else if (current > max) {
            max = current;
            maxOccurs = 1;
        }
    }

    System.out.println("Min is " + min + " and has " + minOccurs + " occurrences");
    System.out.println("Max is " + max + " and has " + maxOccurs + " occurrences");
    // prints: "Min is 1 and has 2 occurrences"
    // prints: "Max is 6 and has 1 occurrences"
}

Answer 4

这是一种将一行由空格分隔的整数解析为int数组的方法：

int[] array = Arrays
            .stream(scanner.nextLine().trim().split("\\s+"))
            .mapToInt(Integer::parseInt)
            .toArray();

Answer 5

这应该为您工作。如果您有任何问题，请告诉我：

import java.util.Scanner;

public class homework2 {

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        System.out.println("Enter the length of the array:");
        int length = s.nextInt();
        System.out.println("Enter the elements of the array:");
        int[] numbers = new int[length];
        for (int i = 0; i < length; i++) {
            numbers[i] = Integer.parseInt(s.next());
        }
        // take first element as max and min
        int max = numbers[0];
        int min = numbers[0];
        // max and min exists so occurs at least once
        int maxOcc = 1;
        int minOcc = 1;
        // start from i = 1; because we assigned the first element before
        for (int i = 1; i < length; i++) {
            int number = numbers[i];
            if (number > max) {
                max = number;
                // start counting again
                maxOcc = 1;
            } else if (number == max) {
                maxOcc++;
            }
            if (number < min) {
                min = number;
                // start counting again
                minOcc = 1;
            } else if (number == min) {
                minOcc++;
            }
        }
        System.out.println("max: " + max);
        System.out.println("maxOcc: " + maxOcc);
        System.out.println("min: " + min);
        System.out.println("minOcc: " + minOcc);
    }
}

Answer 6

这里是您的问题的解决方案。这需要一个带数字和空格的字符串，将它们添加到动态数组（ArrayList）中，并使用min，max函数。不要初始化min和max = 0，因为我们必须从提供的列表中选择一个整数才能开始比较，并且必须是第一个元素[0]。

代码

import java.util.ArrayList;
import java.util.Scanner;
public class homework2 { 
//Using static with functions because of the function calls in the Static 
//main function
//Find the maximum number
public static int max(ArrayList<Integer> array) {
 int max=array.get(0);
    for(int i=0; i<array.size(); i++ ){
      if(array.get(i)>max) 
        max = array.get(i); 
 }
   return max;
 }

//Calculate Maximum number's occurences
public static int maxOccur(int max,ArrayList<Integer> array){
  int maxOccur=0;
   for(int i=0; i<array.size(); i++ )
      if(max==array.get(i)) maxOccur++;
  return maxOccur;
 }

//Find the minimum number
public static int min(ArrayList<Integer> array) {
 int min=array.get(0);
    for(int i=0; i<array.size(); i++ ){
      if(array.get(i)<min) 
        min = array.get(i);
  }  
  return min;
  }

 //Calculate Minimum number's occurences
 public static int minOccur(int min,ArrayList<Integer> array){
  int minOccur=0;
  for(int i=0; i<array.size(); i++ )
      if(min==array.get(i)) minOccur++;
  return minOccur;
  }

 public static void main(String args[]) 
 { 
      int minNum,maxNum; 
      Scanner in = new Scanner(System.in);  
      System.out.print("Enter the numbers separated by a space: ");  
      String number = in.nextLine();

    //Separate the string by spaces in-between
     String[] separatedNums = number.split(" ");
     //Create a dynamic ArrayList to store any amount of integers
    ArrayList<Integer> arrayList = new ArrayList<Integer>();
    //Save the above string in the ArrayList after parsing into integer
     for (String a : separatedNums) 
        arrayList.add(Integer.parseInt(a));
    minNum=min(arrayList);
    maxNum=max(arrayList);
     //Output the results
    System.out.println("Minimum Number="+minNum+" has 
    "+minOccur(minNum,arrayList)+" occurances");
    System.out.println("Maximum Number="+maxNum+" has 
    "+maxOccur(maxNum,arrayList)+" occurances");
     //Close the scanner
      in.close();
  } 
  }

如何找到每个事件的最小值和最大值以及出现次数

6 个答案: