【发布时间】:2015-08-15 10:51:09
【问题描述】:
我使用不维护对尾节点的引用的链表编写了队列的以下实现。当我尝试打印队列时,它只输出头部,即只输出一个节点。错误是什么?提前致谢!
package DataStructures;
import java.util.Scanner;
class Node {
int x;
Node nextNode;
public Node(int x) {
this.x = x;
nextNode = null;
}
}
class Queue {
Node head = null;
int n = 0;
public void enqueue(int x) {
if (n==0){
head = new Node(x);
n++;
return;
}
Node tempHead = head;
while (tempHead != null){
tempHead = tempHead.nextNode;
}
tempHead = new Node(x);
tempHead.nextNode = null;
n++;
}
public int dequeue() {
if (head == null) {
throw new Error("Queue under flow Error!");
} else {
int x = head.x;
head = head.nextNode;
return x;
}
}
public void printTheQueue() {
Node tempNode = head;
System.out.println("hi");
while (tempNode != null){
System.out.print(tempNode.x + " ");
tempNode = tempNode.nextNode;
}
}
}
public class QueueTest {
private static Scanner in = new Scanner(System.in);
public static void main(String[] args) {
Queue queue = new Queue();
while (true){
int x = in.nextInt();
if (x == -1){
break;
} else{
queue.enqueue(x);
}
}
queue.printTheQueue();
}
}
【问题讨论】:
-
当您入队时,
head和您的新节点之间没有任何连接。 -
@RealSkeptic 但我确实有一个对 head 的临时引用并通过它传播以到达最后一个节点。到达最后一个节点后,我将其 nextNode 指向具有数据键 x 的新节点。
标签: java linked-list queue