我正在完成一项任务,我需要在给定模板的情况下创建链接列表。对于每个创建的新节点,我需要打印出更新的列表。但是,到目前为止,我一直难以打印出如何打印链表。任何人都可以弄清楚我做错了什么?我目前只打印出已创建的数字,然后是空格,而不是整个列表。
NumberList.java
import java.util.*;
public class NumberList {
private Node head;
public NumberList() {
}
public void insertAtHead(int x) {
Node newNode = new Node(x);
if (head == null)
head = newNode;
else {
newNode.setNext(head);
head = newNode;
}
}
public void insertAtTail(int x) {
}
public void insertInOrder(int x) {
}
public String toString() {
Node tmp = head;
String result = "";
while (tmp.getNext() != null) {
result += tmp.toString() + " ";
}
return result;
}
//---------------------
// test methods
//---------------------
public static void testInsertAtHead() {
Random r = new Random();
int n = 20;
int range = 1000;
NumberList list = new NumberList();
for (int i=1; i<=n; i++) {
int x = r.nextInt(range);
list.insertAtHead(x);
System.out.println("" + x + ": " + list);
}
}
public static void testInsertAtTail() {
Random r = new Random();
int n = 20;
int range = 1000;
NumberList list = new NumberList();
for (int i=1; i<=n; i++) {
int x = r.nextInt(range);
list.insertAtTail(x);
System.out.println("" + x + ": " + list);
}
}
public static void testInsertInOrder() {
Random r = new Random();
int n = 20;
int range = 1000;
NumberList list = new NumberList();
for (int i=1; i<=n; i++) {
int x = r.nextInt(range);
list.insertInOrder(x);
System.out.println("" + x + ": " + list);
}
}
public static void main(String[] args) {
//testInsertAtHead();
//testInsertAtTail();
testInsertInOrder();
}
}
Node.java
class Node {
private int number;
private Node next;
public Node(int n) {
this.number = n;
this.next = null;
}
public Node getNext() {
return next;
}
public int getNumber() {
return number;
}
public void setNext(Node n) {
if (n == null)
return;
n.setNext(next);
next = n;
}
public String toString() {
return number + "";
}
}
答案 0 :(得分:1)
我认为一旦添加第二个元素,你的toString()
就会无休止地循环。您需要将指针移动到下一个节点:
public String toString() {
Node tmp = head;
String result = "";
while (tmp != null) {
result += tmp.toString() + " ";
tmp = tmp.getNext();
}
return result;
}
我还更新了处理空head
的条件。