Java 中的 IntegerList 帮助

Question

因此，我们在课堂上遵循实验室手册的说明。我能够完成第一步和第二步，我只需要第三步的帮助。

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实验室手册说明：

1. 将方法 void removeFirst(int newVal) 添加到 IntegerList 类，从列表中删除第一次出现的值。如果该值没有出现在列表中，它应该什么都不做（但这不是错误）。删除一个项目不应该改变数组的大小，但请注意，数组值确实需要保持连续，所以当你删除一个值时，你必须将它之后的所有内容向下移动以填满它的空间。还要记住减少跟踪元素数量的变量。

在 IntegerListTest 的菜单中添加一个选项来测试您的新方法。

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整数列表

public class IntegerList
{
    private int count;
    private double totalInt;
    int[] list; //values in the list
    //-------------------------------------------------------
    //create a list of the given size
    //-------------------------------------------------------

    void addElement(int newVal)
    {
        if (count == list.length)
            increaseSize();

        list[count] = newVal;
        count++;
    }

    void removeFirst(int newVal2)
    {
        for (int i = 0; i < list.length-1; i++)
        {
            if (newVal2 == list[i])
            {
                list[list.length] =  (Integer) null;
                list[i] = list [i-1];
            }
        }

    }

    public IntegerList(int size)



    {
        list = new int[size];
        count = 0;

    }

    public void randomize()
    {
        for (int i=0; i<list.length; i++)
            {
            list[i] = (int)(Math.random() * 100) + 1;
            count++;
            }

    }
            public void print()
    {
        for (int i=0; i<count; i++)
            System.out.println(i + ":\t" + list[i]);
    }




private void increaseSize()
{
    int[] temp = new int[list.length * 2];

    for (int lst = 0; lst < list.length; lst++)
        temp[lst] = list[lst];

    list = temp;
}
}

整数列表测试

import java.util.Scanner;
public class IntegerListTest
{
    static IntegerList list = new IntegerList(10);
    static Scanner scan = new Scanner(System.in);

    public static void main(String[] args)
    {
        printMenu();
        int choice = scan.nextInt();
        while (choice != 0)
        {
            dispatch(choice);
            printMenu();
            choice = scan.nextInt();
        }
    }

公共静态无效调度（int选择） {

int loc;
switch(choice)
{
case 0:
    System.out.println("Bye! ") ;
    break;
case 1:
    System.out.println("How big should the list be?");
    int size = scan.nextInt();
    list = new IntegerList(size);
    list.randomize();
    break;
case 2:
    list.print();
    break;
case 3:
    System.out.println("What number would you like to add?");
    int newVal = scan.nextInt();
    list.addElement(newVal);
    break;
case 4:
    System.out.println("What number do you want to remove? (Removes first occurance.)");
    int newVal2 = scan.nextInt();
    list.removeFirst(newVal2);
default:
    System.out.println("Sorry, invalid choice");
}
}


public static void printMenu()
{
    System.out.println("\n Menu ");
    System.out.println(" ====");
    System.out.println("0: Quit");
    System.out.println("1: Create a new list (** do this first!! **)");
    System.out.println("2: Print the list");
    System.out.println("3: Add to the list");
    System.out.println("4: Remove Integer");
    System.out.print("\nEnter your choice: ");
        }
    }

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非常感谢任何帮助。如果你也能解释为什么我可以从中学习，那就太酷了。谢谢！ :D

Answer 1

因此，您可以循环数组以查找与给定值匹配的第一个元素。 如果找到，将其位置保存到索引并将其后的所有元素移动一个位置：

void removeFirst(int newVal2) {
  int index = -1;
  for (int i = 0; i < count; i++) {
     if (index == -1 && newVal2 == list[i]) {
       // element found - save index
       index = i;
     }
     if (index != -1) {
        // this code handles after found case
        if (i == count-1) {
          // zero-out last element
          list[i] = 0;
          count--;
        }
        else {
          // shift value
          list[i] = list[i+1];
        }
     }
  }
}   

Answer 2

public void removeFirst(int val){
    //the position at which the algorithm currently is
    int i = 0;

    boolean found = false;

    //search for the first occurence of val in the array
    for(; i < count && list[i] != val ; i++);
    //i is now the index of the first value equal to val in list

    //if a match was found, the index must be smaller than the array-size
    found = (i < count);

    //shift all elements that are right of the value to the left
    //to leave no empty space. Since the value at i is the searched value  
    //it's left out (i += 1)
    i += 1;
    for(; i < count ; i++)
        list[i - 1] = list[i];//move the current element one to the left

    //check if a match was found and decrement the sizecounter, if one was found
    if(found)
        --count;
}

Answer 3

我真的没有时间给你很多细节，我知道这不是一个真正有效的方法，但你可以做的是总是创建一个“size-1”的新数组，然后记下您要删除的数组元素的索引，然后将该数组元素之前和之后的所有内容复制到您的新数组中，一旦完成所有这些，您就可以将其复制回原始数组，就像这样 oldArray = newArray.