比较冒泡排序
CompareTo bubble sort
我正在尝试使用冒泡排序方法对程序中的值进行排序。我相信我在 organisedRoom 方法中的代码是正确的。但是,当我 运行 代码添加一些客户然后尝试对它们进行排序时,程序崩溃了。如果有人能指出正确的方向,我将不胜感激。
package test;
import java.io.IOException;
import java.util.Scanner;
public class Test {
private class Customer implements Comparable<Customer>{
private String name;
public Customer(String name) {
this.name = name;
}
//Override to stop the program returning memory address as string
@Override
public String toString() {
return name;
}
@Override
public int compareTo(Customer c) {
return name.compareTo(c.name);
}
}
//Array to store customers
public Customer[] customers;
public Scanner input = new Scanner(System.in);
public Test(int nRooms) throws IOException {
customers = new Test.Customer[nRooms];
System.out.println("Welcome to the Summer Tropic Hotel\n");
chooseOption();
}
final JFileChooser fc = new JFileChooser();
// Call new Hotel with int value to allocate array spaces
public static void main(String[] args) throws IOException {
Test t = new Test(11);
}
// New procedure to return User input and point to next correct method
private String chooseOption() throws IOException {
// Set to null, this will take user input
String choice;
//Menu options
System.out.println("This is the Hotel Menu. Please choose from the following options:\n");
System.out.println("A: " + "This will add a new entry\n");
System.out.println("O: " + "View booked rooms, in order of customers name.\n");
System.out.println("X: " + "Exit the program\n");
// Take user input and assign it to choice
choice = input.next();
// Switch case used to return appropriate method
switch (choice.toUpperCase()) {
case "A" :
System.out.println("");
addCustomer();
return this.chooseOption();
case "O" :
System.out.println("");
organisedRoom();
return this.chooseOption();
case "X":
System.exit(0);
}
return choice;
}
// Add a new customer to the Array
public void addCustomer() throws IOException {
// New variable roomNum
int roomNum = 1;
// Loop
do {
// Take user input as room number matching to array index - 1
System.out.println("Please choose a room from 1 to 10");
roomNum = input.nextInt() - 1;
// If room is already booked print this
if (customers[roomNum] != null) {
System.out.println("Room " + roomNum + 1 + " is not free, choose a different one.\n");
this.addCustomer();
}
// Do until array index does not equal to null
} while (customers[roomNum]!= null);
System.out.println("");
// User input added to array as name replacing null (non case-sensetive)
System.out.println("Now enter a name");
customers[roomNum] = new Customer(input.next().toLowerCase());
// Customer (name) added to room (number)
System.out.println(String.format("Customer %s added to room %d\n", customers[roomNum], roomNum + 1));
}
private void organisedRoom() {
boolean flag = true;
Customer temp;
int j;
while (flag) {
flag = false;
for (j = 0; j < customers.length - 1; j++) {
if (customers[j].compareTo(customers[j+1]) < 0) {
temp = customers[j];
customers[j] = customers[j + 1];
customers[j + 1] = temp;
flag = true;
}
}
}
}
}
我认为这是因为数组的初始化将null添加到所有数组索引位置。
堆栈跟踪如下:
Exception in thread "main" java.lang.NullPointerException
at test.Test$Customer.compareTo(Test.java:34)
at test.Test.organisedRoom(Test.java:133)
at test.Test.chooseOption(Test.java:83)
at test.Test.chooseOption(Test.java:79)
at test.Test.chooseOption(Test.java:79)
at test.Test.<init>(Test.java:46)
at test.Test.main(Test.java:55)
Java Result: 1
它失败了,因为您创建了 Customer[]
,它将使用 11 null
个引用进行初始化。如果要对它们进行排序,将比较数组中的 all 个元素。其中引入java.lang.NullPointerException
.
将 Customer
存储在 ArrayList
中。那么你应该可以避免这个错误。
编辑
如果你真的需要尽可能接近你当前的代码。以下将修复您的排序。 (不要将此解决方案用于现实生活中的项目)
private void organisedRoom() {
for (int i = customers.length - 1; i > 0; i--) {
for (int j = 0; j < i; j++) {
if (customers[j + 1] == null) {
continue;
}
if (customers[j] == null ||customers[j + 1].compareTo(customers[j]) < 0) {
Customer temp = customers[j + 1];
customers[j + 1] = customers[j];
customers[j] = temp;
}
}
}
System.out.println("show rooms: " + Arrays.toString(customers));
}
编辑 2
为了保留大部分当前代码,您可以将房间存储在 Customer
实例中(我个人不喜欢)。
// change the constructor of Customer
public Customer(String name, int room) {
this.name = name;
this.room = room;
}
// change the toString() of Customer
public String toString() {
return String.format("customer: %s room: %d", name, room);
}
// store the Customer like
customers[roomNum] = new Customer(input.next().toLowerCase(), roomNum);
您对冒泡排序的实现不正确。它使用嵌套的 for 循环。
for(int i = 0; i < customers.length; i++)
{
for(int j = 1; j < (customers.length - i); j++)
{
if (customers[j-1] > customers[j])
{
temp = customers[j-1];
customers[j-1] = customers[j];
customers[j] = temp;
}
}
}
我正在尝试使用冒泡排序方法对程序中的值进行排序。我相信我在 organisedRoom 方法中的代码是正确的。但是,当我 运行 代码添加一些客户然后尝试对它们进行排序时,程序崩溃了。如果有人能指出正确的方向,我将不胜感激。
package test;
import java.io.IOException;
import java.util.Scanner;
public class Test {
private class Customer implements Comparable<Customer>{
private String name;
public Customer(String name) {
this.name = name;
}
//Override to stop the program returning memory address as string
@Override
public String toString() {
return name;
}
@Override
public int compareTo(Customer c) {
return name.compareTo(c.name);
}
}
//Array to store customers
public Customer[] customers;
public Scanner input = new Scanner(System.in);
public Test(int nRooms) throws IOException {
customers = new Test.Customer[nRooms];
System.out.println("Welcome to the Summer Tropic Hotel\n");
chooseOption();
}
final JFileChooser fc = new JFileChooser();
// Call new Hotel with int value to allocate array spaces
public static void main(String[] args) throws IOException {
Test t = new Test(11);
}
// New procedure to return User input and point to next correct method
private String chooseOption() throws IOException {
// Set to null, this will take user input
String choice;
//Menu options
System.out.println("This is the Hotel Menu. Please choose from the following options:\n");
System.out.println("A: " + "This will add a new entry\n");
System.out.println("O: " + "View booked rooms, in order of customers name.\n");
System.out.println("X: " + "Exit the program\n");
// Take user input and assign it to choice
choice = input.next();
// Switch case used to return appropriate method
switch (choice.toUpperCase()) {
case "A" :
System.out.println("");
addCustomer();
return this.chooseOption();
case "O" :
System.out.println("");
organisedRoom();
return this.chooseOption();
case "X":
System.exit(0);
}
return choice;
}
// Add a new customer to the Array
public void addCustomer() throws IOException {
// New variable roomNum
int roomNum = 1;
// Loop
do {
// Take user input as room number matching to array index - 1
System.out.println("Please choose a room from 1 to 10");
roomNum = input.nextInt() - 1;
// If room is already booked print this
if (customers[roomNum] != null) {
System.out.println("Room " + roomNum + 1 + " is not free, choose a different one.\n");
this.addCustomer();
}
// Do until array index does not equal to null
} while (customers[roomNum]!= null);
System.out.println("");
// User input added to array as name replacing null (non case-sensetive)
System.out.println("Now enter a name");
customers[roomNum] = new Customer(input.next().toLowerCase());
// Customer (name) added to room (number)
System.out.println(String.format("Customer %s added to room %d\n", customers[roomNum], roomNum + 1));
}
private void organisedRoom() {
boolean flag = true;
Customer temp;
int j;
while (flag) {
flag = false;
for (j = 0; j < customers.length - 1; j++) {
if (customers[j].compareTo(customers[j+1]) < 0) {
temp = customers[j];
customers[j] = customers[j + 1];
customers[j + 1] = temp;
flag = true;
}
}
}
}
}
我认为这是因为数组的初始化将null添加到所有数组索引位置。
堆栈跟踪如下:
Exception in thread "main" java.lang.NullPointerException
at test.Test$Customer.compareTo(Test.java:34)
at test.Test.organisedRoom(Test.java:133)
at test.Test.chooseOption(Test.java:83)
at test.Test.chooseOption(Test.java:79)
at test.Test.chooseOption(Test.java:79)
at test.Test.<init>(Test.java:46)
at test.Test.main(Test.java:55)
Java Result: 1
它失败了,因为您创建了 Customer[]
,它将使用 11 null
个引用进行初始化。如果要对它们进行排序,将比较数组中的 all 个元素。其中引入java.lang.NullPointerException
.
将 Customer
存储在 ArrayList
中。那么你应该可以避免这个错误。
编辑
如果你真的需要尽可能接近你当前的代码。以下将修复您的排序。 (不要将此解决方案用于现实生活中的项目)
private void organisedRoom() {
for (int i = customers.length - 1; i > 0; i--) {
for (int j = 0; j < i; j++) {
if (customers[j + 1] == null) {
continue;
}
if (customers[j] == null ||customers[j + 1].compareTo(customers[j]) < 0) {
Customer temp = customers[j + 1];
customers[j + 1] = customers[j];
customers[j] = temp;
}
}
}
System.out.println("show rooms: " + Arrays.toString(customers));
}
编辑 2
为了保留大部分当前代码,您可以将房间存储在 Customer
实例中(我个人不喜欢)。
// change the constructor of Customer
public Customer(String name, int room) {
this.name = name;
this.room = room;
}
// change the toString() of Customer
public String toString() {
return String.format("customer: %s room: %d", name, room);
}
// store the Customer like
customers[roomNum] = new Customer(input.next().toLowerCase(), roomNum);
您对冒泡排序的实现不正确。它使用嵌套的 for 循环。
for(int i = 0; i < customers.length; i++)
{
for(int j = 1; j < (customers.length - i); j++)
{
if (customers[j-1] > customers[j])
{
temp = customers[j-1];
customers[j-1] = customers[j];
customers[j] = temp;
}
}
}