英文:
Bubble sort with swap function having reference as parameter, getting runtime error
问题
我为冒泡排序编写了一个简单的代码，并使用了一个接受对象引用作为参数的交换函数。我遇到了以下错误：
Exception in thread "main" java.util.NoSuchElementException
at java.util.Scanner.throwFor(Scanner.java:862)
at java.util.Scanner.next(Scanner.java:1485)
at java.util.Scanner.nextInt(Scanner.java:2117)
at java.util.Scanner.nextInt(Scanner.java:2076)
at Main.main(Main.java:6)
我的代码：
import java.util.Scanner;
public class Bubble {
public static void main(String[] args){
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int[] arr = new int[n];
for(int i = 0; i < n ; i++){
arr[i] = scan.nextInt();
}
bubbleSort(arr);
for(int i = 0; i < n ; i++){
System.out.print(arr[i] + " ");
}
}
static int[] bubbleSort(int[] arr){
int swapped = 1;
while(swapped != 0){
swapped = 0;
for(int i = 1; i < arr.length; i++){
if(arr[i - 1] > arr[i]){
Example a = new Example(arr[i - 1]);
Example b = new Example(arr[i]);
swap(a, b);
swapped++;
}
}
}
return arr;
}
static void swap(Example a, Example b){
int temp = a.a;
a.a = b.a;
b.a = temp;
}
}
class Example{
int a ;
public Example(int a){
this.a = a;
}
}
我该如何捕获这些错误，为什么会出现这些错误？
英文:
I wrote a simple code for bubble sort and used a swap function which accepts reference of objects as its parameter.
I am getting the following error:
Exception in thread "main" java.util.NoSuchElementException
at java.util.Scanner.throwFor(Scanner.java:862)
at java.util.Scanner.next(Scanner.java:1485)
at java.util.Scanner.nextInt(Scanner.java:2117)
at java.util.Scanner.nextInt(Scanner.java:2076)
at Main.main(Main.java:6)
My code:
import java.util.Scanner;
public class Bubble {
public static void main(String[] args){
Scanner scan = new Scanner(System.in);
int n= scan.nextInt();
int[] arr = new int[n];
for(int i = 0; i<n ; i++){
arr[i] = scan.nextInt();
}
bubbleSort(arr);
for(int i = 0; i<n ; i++){
System.out.print(arr[i]+"");
}
}
static int[] bubbleSort(int[] arr){
int swapped = 1;
while(swapped!= 0){
swapped = 0;
for(int i=1; i< arr.length; i++){
if(arr[i-1]>arr[i]){
Example a = new Example(arr[i-1]);
Example b = new Example(arr[i]);
swap(a,b);
swapped++;
}
}
}
return arr;
}
static void swap(Example a, Example b){
Example temp = new Example(0) ;
temp.a = a.a;
a.a = b.a;
b.a= temp.a;
}
}
class Example{
int a ;
public Example(int a){
this.a = a;
}
}
How can I catch these errors and why am I getting these?
答案1
得分: 0
你对 Example 的操作是不起作用的。当你执行：
Example a = new Example(arr[i-1]);
你的 Example a 只包含了 arr[i-1] 的值。对 a.a 的修改不会影响你的数组。所以你的数组永远不会变成有序的，而且会陷入无限循环。
如果你想要一个交换数组中元素的方法，最简单的方式是将数组传递给方法。
static void swap(int[] arr, int i, int j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
然后你可以使用以下方法来交换元素：
swap(arr, i-1, i);
英文:
What you're doing with Example will not work. When you do
Example a = new Example(arr[i-1]);
Your Example a only contains the value of arr[i-1]. Altering a.a has no effect on your array. So your array never becomes sorted and you have an infinite loop.
If you want a method to swap elements in your array, the easiest way is to pass the array to the method.
static void swap(int[] arr, i, j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
Then you can swap elements with
swap(arr, i-1, i);
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