Java 从单向链表打印节点
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printing nodes from a singly-linked list
提问by user2326847
I made a node class which is a linked list class. Is there any way I can print out elements in this list ? I made my print()method but it only returns the first element which is 21. How do I iterate through that list ?
我做了一个节点类，它是一个链表类。有什么办法可以打印出这个列表中的元素吗？我创建了我的print()方法，但它只返回第一个元素，即 21。如何遍历该列表？
public class ListNode {
private int item;
private ListNode next;
public ListNode(int item, ListNode next){
this.item = item;
this.next = next;
}
public ListNode(int item){
this(item, null);
}
public int print(){
return item;
}
public static void main(String[] args) {
ListNode list = new ListNode(21, new ListNode(5, new ListNode(19, null)));
System.out.println(list.print());
}
}
}
采纳答案by ValarDohaeris
public String toString() {
String result = item + " ";
if (next != null) {
result += next.toString();
}
return result;
}
And then you can simply do
然后你可以简单地做
System.out.println(list.toString());
(I renamed your function from printto toStringto give a more accurate description of what it does)
（我将您的函数重命名为printtotoString以更准确地描述它的作用）
回答by John B
Consider creating a printall
考虑创建一个 printall
public void printAll(){
System.out.println(item);
if (next != null){
next.printAll();
}
}
回答by Vladimir
Your current implementation doesn't print anything: it simply returns item. More appropriate implementation would look like:
您当前的实现不打印任何内容：它只是返回项目。更合适的实现如下所示：
public void print() {
System.out.println(item);
}
You can then use recursion to print all items:
然后您可以使用递归打印所有项目：
public void printAll() {
print();
if (next != null) {
System.out.println("; ");
next.printAll();
}
}
回答by goodies
Your print()function is returning only a single item that's why it is printing only 21.
您的print()函数只返回一个项目，这就是为什么它只打印 21。
Call recursively to print all the values until next != NULL
递归调用以打印所有值，直到 next != NULL
回答by jayrobin
Calling list.print()will only ever return the value of the head (21) - you are never making any reference or call to the next node: next.
调用list.print()将只返回头（21）的价值-你永远不会做任何参考或呼叫到下一个节点：next。
Personally, I would remove the print() method and instead override toString():
就个人而言，我会删除 print() 方法，而是覆盖 toString()：
@override
public String toString(){
return item + "\n" + next;
}
I guess you probably don't want the null tail printed, so this is probably nicer:
我想你可能不希望打印空尾，所以这可能更好：
@override
public String toString(){
if(next) {
return item + "\n" + next;
} else {
return item + "\n";
}
}
Then in, main:
然后在，主要：
public static void main(String[] args) {
ListNode list = new ListNode(21, new ListNode(5, new ListNode(19, null)));
System.out.println(list);
}
回答by John Snow
You can use a foreach loop:
您可以使用 foreach 循环：
List<ListNode> theList = new LinkedList<ListNode>();
//add stuff to the list
for(ListNode n:theList)
System.out.println(n.print();
THis will iterate over the list and return the next object, on this object we call the print()method
这将遍历列表并返回下一个对象，在这个对象上我们调用print()方法
