java - 如何从输入的一组数字中获取最大值和最小值？

Question

以下是我到目前为止的内容：

我不知道如何排除 0 作为最小数字。作业要求 0 作为退出编号，因此我需要在 min 字符串中显示除 0 之外的最小数字。有任何想法吗？

int min, max;

Scanner s = new Scanner(System.in);
System.out.print("Enter a Value: ");
int val = s.nextInt();
min = max = val;

while (val != 0) {
  System.out.print("Enter a Value: ");
  val = s.nextInt();
  if (val < min) {
      min = val;
  }
  if (val > max) {
     max = val;
  }
};
System.out.println("Min: " + min);
System.out.println("Max: " + max);

Answer 1

这是一个可能的解决方案：

public class NumInput {
  public static void main(String [] args) {
    int min = Integer.MAX_VALUE;
    int max = Integer.MIN_VALUE;

    Scanner s = new Scanner(System.in);
    while (true) {
      System.out.print("Enter a Value: ");
      int val = s.nextInt();

      if (val == 0) {
          break;
      }
      if (val < min) {
          min = val;
      }
      if (val > max) {
         max = val;
      }
    }

    System.out.println("min: " + min);
    System.out.println("max: " + max);
  }
}

（不确定是否使用 int 或双重思考）

Answer 2

您只需要像这样跟踪最大值：

int maxValue = 0;

然后当您遍历这些数字时，如果它大于 maxValue，则继续将 maxValue 设置为下一个值：

if (value > maxValue) {
    maxValue = value;
}

以相反的方向重复 minValue。

Answer 3

这个比较好

public class Main {

    public static void main(String[] args) {

        System.out.print("Enter numbers: ");
        Scanner input = new Scanner(System.in);

        double max = Double.MIN_VALUE;
        double min = Double.MAX_VALUE;

        while (true) {

            if ( !input.hasNextDouble())
                break;

            Double num = input.nextDouble();

            min = Math.min(min, num);
            max = Math.max(max, num);

        }
        System.out.println("Max is: " + max);
        System.out.println("Min is: " + min);
    }
}

Answer 4

//for excluding zero
public class SmallestInt {

    public static void main(String[] args) {

        Scanner input= new Scanner(System.in);

        System.out.println("enter number");
        int val=input.nextInt();
        int min=val;

        //String notNull;

        while(input.hasNextInt()==true)
        {
            val=input.nextInt();
            if(val<min)
                min=val;
        }
        System.out.println("min is: "+min);
    }
}

Answer 5

这就是我所做的，它可以尝试并尝试使用它。它计算总、平均、最小值和最大值。

public static void main(String[] args) {
   int[] score= {56,90,89,99,59,67};
   double avg;
   int sum=0;
   int maxValue=0;
   int minValue=100;

   for(int i=0;i<6;i++){
       sum=sum+score[i];
   if(score[i]<minValue){
    minValue=score[i];
   }
   if(score[i]>maxValue){
    maxValue=score[i];
   }
   }
   avg=sum/6.0;
System.out.print("Max: "+maxValue+"," +" Min: "+minValue+","+" Avarage: "+avg+","+" Sum: "+sum);}

 }

Answer 6

在这里您需要跳过 int 0 ，如下所示：

val = s.nextInt();
  if ((val < min) && (val!=0)) {
      min = val;
  }

Answer 7

我试图通过处理用户输入异常来优化解决方案。

public class Solution {
    private static Integer TERMINATION_VALUE = 0;
    public static void main(String[] args) {
        Integer value = null;
        Integer minimum = Integer.MAX_VALUE;
        Integer maximum = Integer.MIN_VALUE;
        Scanner scanner = new Scanner(System.in);
        while (value != TERMINATION_VALUE) {
            Boolean inputValid = Boolean.TRUE;
            try {
                System.out.print("Enter a value: ");
                value = scanner.nextInt();
            } catch (InputMismatchException e) {
                System.out.println("Value must be greater or equal to " + Integer.MIN_VALUE + " and less or equals to " + Integer.MAX_VALUE );
                inputValid = Boolean.FALSE;
                scanner.next();
            }
            if(Boolean.TRUE.equals(inputValid)){
                minimum = Math.min(minimum, value);
                maximum = Math.max(maximum, value);
            }
        }
        if(TERMINATION_VALUE.equals(minimum) || TERMINATION_VALUE.equals(maximum)){
            System.out.println("There is not any valid input.");
        } else{
            System.out.println("Minimum: " + minimum);
            System.out.println("Maximum: " + maximum);
        }
        scanner.close();
    }
}

Answer 8

System.out.print("Enter a Value: ");
val = s.nextInt();

这一行放在最后。整个代码如下：-

public static void main(String[] args){
    int min, max;

    Scanner s = new Scanner(System.in);
    System.out.print("Enter a Value: ");
    int val = s.nextInt();
    min = max = val;

    while (val != 0) {
        if (val < min) {
            min = val;
        }
        if (val > max) {
            max = val;
        }
        System.out.print("Enter a Value: ");
        val = s.nextInt();

    }
    System.out.println("Min: " + min);
    System.out.println("Max: " + max);
}