数组变量输入帮助[关闭]

发布于 2024-10-03 08:07:27 字数 1099 浏览 6 评论 0原文

首先，我必须创建一个长度为 n 的数组，输入变量来填充该数组，然后在数组位置 k 处，我必须将所有数组位置 k 及向上推 1，并将 x 的值放入数组位置 k 中。但如果 k = n，则将 x 放入 n+1 中。

首先，我在使 k 和 x 的整数工作时遇到问题。由于某种原因，代码将 n 设置为第一个输入，并在设置 n 后立即将 k 和 x 设置为与 n 相同。

其次，我在将数组扩展到 n+1 时遇到问题。我知道不只是给我答案，但我需要一些前进的方向。

import java.util.Scanner;
public class hw2
{
   public static void main(String[] args) 
   {
    Scanner scan = new Scanner(System.in);
    int n = scan.nextInt();
    int k = scan.nextInt();
    int x = scan.nextInt();
    int[] a = new int[n];
    for(int i = 0; i<n; i++)
    {
     a[i] = scan.nextInt();
    }
    n++;
    final int LENGTH = a.length - 1;
    for(int j=LENGTH; j>k; j--)
    {
     a[j] = a[j-1];
    }
    a[k] = x;
    for(int h = 0; h < n; h++)
    {
     System.out.println("location " + h + " is " + a[h]);    
    }

   }
}

nkxa[0]... a[n-1] 的样本输入

5  3  7  2  3  5  11  13

分别为。

使用该输入运行代码后，n 应该 = 6 并且数组应该

a[0] = 2
a[1] = 3
a[2] = 5
a[3] = 7
a[4] = 11
a[5] = 13

编辑：我读到的问题完全错误。底部的提示说“假设数组的大小至少为 n+1” ...

原文

First of all, I have to create an array of length n, input variables to fill that array, then at array location k, I have to push all arrays location k and up by 1 and put the value of x into array location k. But if k = n, then put x in n+1.

First of all, I am having a problem making the ints for k and x work. For some reason the code sets n to the first input and sets k and x to the same as n immediately when n is set.

Secondly, I am having trouble extending the array to n+1. I know don't just give me the answer but I need some direction on where to go.

import java.util.Scanner;
public class hw2
{
   public static void main(String[] args) 
   {
    Scanner scan = new Scanner(System.in);
    int n = scan.nextInt();
    int k = scan.nextInt();
    int x = scan.nextInt();
    int[] a = new int[n];
    for(int i = 0; i<n; i++)
    {
     a[i] = scan.nextInt();
    }
    n++;
    final int LENGTH = a.length - 1;
    for(int j=LENGTH; j>k; j--)
    {
     a[j] = a[j-1];
    }
    a[k] = x;
    for(int h = 0; h < n; h++)
    {
     System.out.println("location " + h + " is " + a[h]);    
    }

   }
}

sample input for n k x a[0]... a[n-1] is

5  3  7  2  3  5  11  13

respectively.

After the code runs with that input, n should = 6 and the array should be

a[0] = 2
a[1] = 3
a[2] = 5
a[3] = 7
a[4] = 11
a[5] = 13

EDIT: I read the problem completely wrong. A hint at the bottom said "Assume that the array is of at least size n+1"
...

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香草可樂 2024-10-10 08:07:27

看起来一切都很好，包括这个：

import java.util.Scanner;
public class hw2
{
   public static void main(String[] args) 
   {
    Scanner scan = new Scanner(System.in);
    int n = scan.nextInt();
    int k = scan.nextInt();
    int x = scan.nextInt();
    int[] a = new int[n];
    for(int i = 0; i<n; i++)
    {
     a[i] = scan.nextInt();
    }

之后就变得有点粗略了。这个问题可以有几种解释......

然后在数组位置k处，输入
变量 x。但如果 k = n，则将 x
在n+1中

对我来说，这意味着“覆盖”位置 k 中的值，除非 k == n。至于这几种方式——这取决于你的作业的措辞方式。如果数组的起始大小并不重要，那么您可以从数组大小 n+1 （或 n+someArbitraryValue）而不是 n 开始。如果您必须根据 k==n 检查将大小保持在 n 或 n+1，那么当 k==n 时，您需要设置一个大小为 n+1 的新数组并将 a 中的值复制到其中。

希望这会有所帮助，但又不会放弃太多。

It looks like everything including this is fine:

import java.util.Scanner;
public class hw2
{
   public static void main(String[] args) 
   {
    Scanner scan = new Scanner(System.in);
    int n = scan.nextInt();
    int k = scan.nextInt();
    int x = scan.nextInt();
    int[] a = new int[n];
    for(int i = 0; i<n; i++)
    {
     a[i] = scan.nextInt();
    }

After that is gets a little sketchy. The problem could be interpreted a couple ways...

then at array location k, input the
variable x. But if k = n, then put x
in n+1

To me, this means "overwrite" the value in location k unless k == n. As to the couple ways - it depends on how your assignment is worded. If it doesn't matter what size the array is to start with then you could start with your array size being n+1 (or n+someArbitraryValue) instead of n. If you have to keep the size at n or n+1 depending on the k==n check then when k==n you need to setup a new array with size n+1 and copy the values from a into it.

Hope this helps without giving too much away.

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妄司 2024-10-10 08:07:27

很难知道你在问什么，因为你的代码与你写的不一致。据我了解，n 元素数组已经填满。在这种情况下，您不能只在末尾附加一个元素，您需要创建一个新数组并将所有数据从旧数组复制到新数组。

它的工作原理如下：

int[] oldArr = new int[n]; //old array, full
int[] newArr = new int[2*n];//new array
for(i=0;i<n;i++){
  new[i] = old[i];
}

然后像平常一样将 k 放入 new[n+1] 中。请注意，如果数组不为空，则只需在 n 处插入 x 即可执行上述操作。

It's hard to know just what you're asking because your code is inconsistent with what you wrote. From what I understand the n element array is already filled to capacity. In this case you can't just append an element at the end, you need to make a new array and copy all of the data from the old array to the new array.

It would work something like this:

int[] oldArr = new int[n]; //old array, full
int[] newArr = new int[2*n];//new array
for(i=0;i<n;i++){
  new[i] = old[i];
}

Then just put k in at new[n+1] like you would normally. Note that if the array were not empty you would only need to do the above if you were inserting x at n.

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