我遇到的问题如下：我创建了两个代表船只停靠空间的数组。第一个数组发送对象保存在数组中，如果没有空格，则将其添加到等待列表数组中。 但是当我从第一个数组中删除一个对象时，等待列表数组中的对象不会被删除并添加。
码头可容纳三种尺寸的船;货物，集装箱和超级集装箱。行由5个小型和3个中型和2个大型组成。货船（小）可以在任何可用空间停泊。集装箱船（中型）可以在中等空间和大型泊位停泊，但不能在狭小的空间内停泊。超级容器只能适应大空间。
因此，如果我输入shipName3和Super-Container，并且已经有两个Super-Container，它会添加到等待列表中，但是当我从Dock中删除一个Super-Container时，它不会从等待列表中删除一个把它添加到码头你能帮忙吗？这是我的dock方法：
import java.util.*;
public class Main {
static Scanner scan = new Scanner(System.in);
private static Ship[] dock1 = new Ship[10];
private static Ship[] waitingList = new Ship[10];
public static void main(String[] args) {
menu();
}
public static void menu() {
Scanner scan = new Scanner(System.in);
while (true) {
System.out.println("Choose an option: 1-3");
System.out.println("1. Dock");
System.out.println("2. Undock");
System.out.println("3. Status");
int menu = scan.nextInt();
switch (menu) {
case 1:
System.out.println("1. Dock");
dock();
break;
case 2:
System.out.println("2. Undock");
undock();
break;
case 3:
System.out.println("3. Status");
printDock();
printWaitingList();
break;
case 4:
System.out.println("4. Exit");
System.exit(0);
default:
System.out.println("No such option");
break;
}
}
}
public static void dock() {
System.out.println("Enter ship's name: ");
String name = scan.nextLine();
System.out.println("Enter ship's size: ");
String size = scan.nextLine();
System.out.println("Enter the ships dock:");
//Check if the dock number is valid
int i = Integer.valueOf(scan.nextLine());
if (i >= 0 && i < 10 && dock1[i] == null) {
int c = 0;
int co = 0;
int sco = 0;
for (int j = 0; j < dock1.length; j++) {
if (dock1[j] != null && dock1[j].getShipSize().equals("Cargo")) {
c++;
}
if (dock1[j] != null && dock1[j].getShipSize().equals("Container")) {
co++;
}
if (dock1[j] != null && dock1[j].getShipSize().equals("Super-Container")) {
sco++;
}
}
if (c < 10 && co < 5 && sco < 2) {
//Add ship to the dock
dock1[i] = new Ship(name, size);
System.out.println("Enough space you can dock");
System.out.println("Ship has been docked");
} else {
System.out.println("You cannot dock");
waitingList(name,size);
}
} else {
System.out.println("Couldn't dock");
waitingList(name, size);
}
}
public static void undock() {
System.out.println("Status of ships: ");
printDock();
System.out.println("Enter ship's name to undock: ");
String name = scan.nextLine();
System.out.println("Enter ship's size to undock: ");
String size = scan.nextLine();
for (int i = 0; i < dock1.length; i++) {
if (dock1[i] != null && dock1[i].getShipName().equals(name)) {
dock1[i] = null;
System.out.println("Ship removed");
//break;
///HERE CHECK IF SHIP IN DOCK
for (int j = 0; j < dock1.length; j++) {
if (dock1[j] == null) {
//Add ship to the dock
dock1[j] = new Ship(name, size);
System.out.println("Move ship from waiting list to dock 1");
break;
} else {
System.out.println("No space in dock1");
return;
}
}
} else {
System.out.println("Ship not docked here");
break;
}
}
}
public static void waitingList(String name, String size){
System.out.println("Dock 1 is full, ship will try to be added to Waiting List");
for (int i = 0; i < dock1.length; i++) { //waitingList?
if (waitingList[i] == null) {
//Add ship to the dock
waitingList[i] = new Ship(name, size);
System.out.println("Enough space added to waiting list");
break;
} else {
System.out.println("No space on waiting list, ship turned away");
return;
}
}
}
public static void printDock() {
System.out.println("Docks:");
for (int i = 0; i < dock1.length; i++) {
if (dock1[i] == null) {
System.out.println("Dock " + i + " is empty");
} else {
System.out.println("Dock " + i + ": " + dock1[i].getShipName() + " " + dock1[i].getShipSize());
}
}
}
private static void printWaitingList() {
System.out.println("Waiting List:");
for (int i = 0; i < waitingList.length; i++) {
if (waitingList[i] == null) {
System.out.println("Dock " + i + " is empty");
} else {
System.out.println("Dock " + i + ": " + waitingList[i].getShipName() + " " + waitingList[i].getShipSize());
}
}
}
}
船舶类
public class Ship {
private String shipName;
private String shipSize;
public String getShipName() {
return shipName;
}
public void setShipName(String shipName) {
this.shipName = shipName;
}
public String getShipSize() {
return shipSize;
}
public void setShipSize(String shipSize) {
this.shipSize = shipSize;
}
public Ship(String shipName, String shipSize) {
this.shipName = shipName;
this.shipSize = shipSize;
}
}
我尝试了解开方法。
答案 0 :(得分：1)
请检查您的取消停靠方法，因为您要从 dock1 数组中删除，然后再将相同的对象添加到 dock1 数组中没有代码从 waitingList 数组中删除对象只需在解除对接方法中更新循环，如下所述
for (int i = 0; i < dock1.length; i++) {
if (dock1[i] != null && dock1[i].getShipName().equals(name)) {
dock1[i] = null;
System.out.println("Ship removed");
// break;
/// HERE CHECK IF SHIP IN DOCK
for (int j = 0; j < waitingList.length; j++) {
if (dock1[i] == null) {
// Add ship to the dock
dock1[i] = new Ship(waitingList[j].getShipName(), waitingList[j].getShipSize());
System.out.println("Move ship from waiting list to dock 1");
waitingList[j]=null;
break;
} else {
System.out.println("No space in dock1");
return;
}
}
} else {
System.out.println("Ship not docked here");
break;
}
}
我已经更改了内循环迭代，因此它将循环到 waitingList 数组的大小，因为我们需要从 waitingList aray中删除对象并添加到 dock1 < / strong> array.Also我运行时从 waitingList 数组中获取 shipname 和 shipize ，以便它将从 waitingList <添加对象/ strong>到 dock1 数组。我已经在我的机器上测试了代码，希望它可以帮助你。
答案 1 :(得分：0)
在您的取消停靠方法中，您可以通过将其设置为null来移除dock1数组中第i个位置的发货。这没关系，但是你有一个for循环，它遍历dock1中的所有船只，寻找一个null值的船。您知道刚删除的那个是null，因为您只需将其设置为该值。然后你将它设置为一艘新船，它具有你移除的船的确切名称和大小（基本上将船返回）。相反，您希望for循环遍历您的等待船只列表，以找到与您刚刚移除的船舶空间相匹配的船舶。
替换：
for (int j = 0; j < dock1.length; j++) {
if (dock1[j] == null) {
//Add ship to the dock
dock1[j] = new Ship(name, size);
System.out.println("Move ship from waiting list to dock 1");
break;
} else {
System.out.println("No space in dock1");
return;
}
}
与
for (int j = 0; j < waitingList.length; j++) {
if (waitingList[j].getShipSize() <= size) {
//Add ship to the dock
dock1[i] = waitingList[j];
System.out.println("Move ship from waiting list to dock 1");
return;
}
}
System.out.println("No space in dock1");
return;