18

我的代码没有给出错误，但是它没有显示最小值和最大值。代码是：

Scanner input = new Scanner(System.in);

int array[] = new int[10];

System.out.println("Enter the numbers now.");

for (int i = 0; i < array.length; i++) {
    int next = input.nextInt();
    // sentineil that will stop loop when 999 is entered
    if (next == 999) {
        break;
    }
    array[i] = next;
    // get biggest number
    getMaxValue(array);
    // get smallest number
    getMinValue(array);
}

System.out.println("These are the numbers you have entered.");
printArray(array);


// getting the maximum value
public static int getMaxValue(int[] array) {
    int maxValue = array[0];
    for (int i = 1; i < array.length; i++) {
        if (array[i] > maxValue) {
            maxValue = array[i];
        }
    }
    return maxValue;
}

// getting the miniumum value
public static int getMinValue(int[] array) {
    int minValue = array[0];
    for (int i = 1; i < array.length; i++) {
        if (array[i] < minValue) {
            minValue = array[i];
        }
    }
    return minValue;
}

//this method prints the elements in an array......
//if this case is true, then that's enough to prove to you that the user input has  //been stored in an array!!!!!!!
public static void printArray(int arr[]) {
    int n = arr.length;

    for (int i = 0; i < n; i++) {
        System.out.print(arr[i] + " ");
    }
}

我需要一个 system.out.println() 来显示它，还是应该返回工作？

4

13 回答 13

18 
getMaxValue(array);
// get smallest number
getMinValue(array);

您正在调用方法但不使用返回的值。

System.out.println(getMaxValue(array));
System.out.println(getMinValue(array));
于 2013-08-30T05:21:16.920 回答
16

你也可以试试这个，如果你不想用你的方法这样做。

    Arrays.sort(arr);
    System.out.println("Min value "+arr[0]);
    System.out.println("Max value "+arr[arr.length-1]);
于 2013-08-30T05:44:52.813 回答
6

恕我直言，最简单的解决方案之一是： -

//MIN NUMBER
Collections.sort(listOfNumbers);
listOfNumbers.get(0);

//MAX NUMBER
Collections.sort(listOfNumbers);
Collections.reverse(listOfNumbers);
listOfNumbers.get(0);
于 2014-12-19T13:16:03.940 回答
6

这是在数组中查找最小值和最大值的工作代码。希望对您有所帮助：

 import java.util.Random;
 import java.util.Scanner;
 public class FindMin {
    public static void main(String[] args){
        System.out.println("Main Method Started");
        Scanner in = new Scanner(System.in);
        System.out.println("Enter the size of the arr");
        int size = in.nextInt();
        System.out.println("Enter the maximum value of the arr");
        int max = in.nextInt();
        int [] arr  = initializeArr(max, size);
        print(arr);
        findMinMax(arr);
        System.out.println("Main Method Ended");
    }
    public static void print(int[] arr){
        for(int val:arr){
            System.out.print(val + " ");
        }
        System.out.println();
    }
    public static int[] initializeArr(int max,int size){
        Random random = new Random();
        int [] arr = new int[size];
        for(int ii=0;ii<arr.length;ii++){
            arr[ii]=random.nextInt(max);
        }
        return arr;
    }
    public static void findMinMax(int[] arr){
        int min=arr[0];
        int max=arr[0];
        for(int ii=0;ii<arr.length;ii++){
            if(arr[ii]<min){
                min=arr[ii];
            }
            else if(arr[ii]>max){
                max=arr[ii];
            }
        }
        System.out.println("The minimum in the arr::"+min);
        System.out.println("The maximum in the arr::"+max);
    }
}
于 2015-02-04T03:12:12.067 回答
3

您只需丢弃 Min/Max 值：

  // get biggest number
  getMaxValue(array); // <- getMaxValue returns value, which is ignored
  // get smallest number
  getMinValue(array); // <- getMinValue returns value, which is ignored as well

你可以做类似的事情

  ... 
  array[i] = next;

  System.out.print("Max value = ");
  System.out.println(getMaxValue(array)); // <- Print out getMaxValue value

  System.out.print("Min value = ");
  System.out.println(getMinValue(array)); // <- Print out getMinValue value

  ...
于 2013-08-30T05:23:35.023 回答
3

你在这里犯了两个错误。1.在数组初始化完成之前
调用方法。 2.不存储方法返回的返回值。 所以试试这段代码getMaxValue(),getMinValue()
getMaxValue(),getMinValue()

   for (int i = 0 ; i < array.length; i++ ) 
  {
       int next = input.nextInt();
       // sentineil that will stop loop when 999 is entered
       if (next == 999)
       break;
       array[i] = next;
  }
  // get biggest number
  int maxValue = getMaxValue(array);
  System.out.println(maxValue );

  // get smallest number
  int minValue = getMinValue(array);
  System.out.println(minValue);
于 2013-08-30T05:24:19.133 回答
3

一行中数组的求和、最大值和最小值

 public static void getMinMaxByArraysMethods(int[] givenArray){

       //Sum of Array in One Line 
       long sumofArray =  Arrays.stream(givenArray).sum();

       //get Minimum Value in an array in One Line
        int minimumValue = Arrays.stream(givenArray).min().getAsInt();

        //Get Maximum Value of an Array in One Line
        int MaxmumValue =  Arrays.stream(givenArray).max().getAsInt();

    }
于 2019-07-28T00:42:06.173 回答
1

您的最大最小方法是正确的

但是您不会将 int 打印到控制台！

和......也许更好的位置变化（最大，最小）方法

现在（最大，最小）方法在循环中。它不需要..只需要一个电话

我建议更改此代码

    for (int i = 0 ; i < array.length; i++ ) {
       int next = input.nextInt();
       // sentineil that will stop loop when 999 is entered
       if (next == 999)
       break;
       array[i] = next;
}
System.out.println("max Value : " + getMaxValue(array));
System.out.println("min Value : " + getMinValue(array));
System.out.println("These are the numbers you have entered.");
printArray(array);
于 2013-08-30T05:25:17.907 回答
1

是的，您需要使用System.out.println. 但是每次他们输入一个值时你都会得到最小值和最大值，并且如果它们提前中断，就不会跟踪元素的数量。

尝试：

for (int i = 0 ; i < array.length; i++ ) {
       int next = input.nextInt();
       // sentineil that will stop loop when 999 is entered
       if (next == 999)
           break;

       array[i] = next;
 }
 int length = i;
 // get biggest number
 int large = getMaxValue(array, length);
 // get smallest number
 int small = getMinValue(array, length);

 // actually print
 System.out.println( "Max: " + large + " Min: " + small );

然后你必须将长度传递给方法来确定最小值和最大值并打印。如果您不这样做，其余字段将为 0，并且可能会弄乱正确的最小值和最大值。

于 2013-08-30T05:32:38.810 回答
1

在这里，您没有打印最大值和最小值。在 getMaxVal 和 getMin val 方法中或在调用之后打印最大值和最小值。这是输出。

Enter the numbers now.
5
Max: 5
Min: 0
3
Max: 5
Min: 0
7
Max: 7
Min: 0
3
Max: 7
Min: 0
90
Max: 90
Min: 0
43
Max: 90
Min: 0
100
Max: 100
Min: 0
45
Max: 100
Min: 0
23
Max: 100
Min: 0
22
Max: 100
Min: 3
These are the numbers you have entered.
5 3 7 3 90 43 100 45 23 22

此外，当您声明一个数组时，它最初全为 0。

于 2013-08-30T05:50:20.837 回答
1 
import java.util.*;
class Maxmin
{
    public static void main(String args[])
    {
        int[] arr = new int[10];
        Scanner in = new Scanner(System.in);
        int i, min=0, max=0;
        for(i=0; i<=arr.length; i++)
        {
            System.out.print("Enter any number: ");
            arr[i] = in.nextInt();          
        }
        min = arr[0];
        for(i=0; i<=9; i++)
        {
            if(arr[i] > max)
            {
                max = arr[i];
            }
            if(arr[i] < min)
            {
                min = arr[i];
            }
        }
        System.out.println("Maximum is: " + max);
        System.out.println("Minimum is: " + min);
    }
}
于 2016-01-07T07:30:03.797 回答
1

//在java中不排序的情况下查找数组中的最大值和最小值

import java.util.Scanner;
import java.util.*;
public class MaxMin_WoutSort {
 public static void main(String args[])
   {
      int n,max=Integer.MIN_VALUE,min=Integer.MAX_VALUE;
      System.out.println("Enter the number of elements: ");
      Scanner sc = new Scanner(System.in);

      int[] arr = new int[sc.nextInt()]; //U can't say static or dynamic.
                                         //UnWrapping object sc to int value;sc.nextInt()
      System.out.println("Enter the elements: ");
      for(int i=0;i<arr.length;i++)      //Loop for entering values in array
      {
          int next = sc.nextInt();
          arr[i] = next;
      }
      for(int j=0;j<arr.length;j++)
     {
          if(arr[j]>max)               //Maximum Condition
            max = arr[j];
          else if(arr[j]<min)         //Minimum Condition
              min = arr[j];
     }
     System.out.println("Highest Value in array: " +max);
     System.out.println("Smallest Value in array: "+min);

 }
}
于 2017-01-15T09:06:02.473 回答
0

我已经更新了您的相同代码，请将代码与您的原始代码进行比较：

public class Help {

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

    int array[] = new int[10];

    System.out.println("Enter the numbers now.");

    for (int i = 0; i < array.length; i++) {
        int next = input.nextInt();
        // sentineil that will stop loop when 999 is entered
        if (next == 999) {
            break;
        }
        array[i] = next;
    }

    System.out.println("These are the numbers you have entered.");
    printArray(array);

    // get biggest number
    System.out.println("Maximum: "+getMaxValue(array));
    // get smallest number
    System.out.println("Minimum: "+getMinValue(array));
}

// getting the maximum value
public static int getMaxValue(int[] array) {
    int maxValue = array[0];
    for (int i = 1; i < array.length; i++) {
        if (array[i] > maxValue) {
            maxValue = array[i];
        }
    }
    return maxValue;
}

// getting the miniumum value
public static int getMinValue(int[] array) {
    int minValue = array[0];
    for (int i = 1; i < array.length; i++) {
        if (array[i] < minValue) {
            minValue = array[i];
        }
    }
    return minValue;
}

//this method prints the elements in an array......
//if this case is true, then that's enough to prove to you that the user input has  //been stored in an array!!!!!!!
public static void printArray(int arr[]) {
    int n = arr.length;

    for (int i = 0; i < n; i++) {
        System.out.print(arr[i] + " ");
    }
}
}
于 2017-07-13T08:58:47.133 回答