Java singly linked list implementation code

The following is a method shared by the editor to write a singly linked list using java. If you have any questions, please leave me a message.

First define a Node class

public class Node {
protected Node next; //指针域
public int data;//数据域
 
public Node( int data) {
this. data = data;
}
//显示此节点
public void display() {
System. out.print( data + " ");
}
}

Next define a singly linked list and implement related methods:

public class LinkList {
public Node first; // 定义一个头结点
private int pos = 0;// 节点的位置
public LinkList() {
this.first = null;
}
// 插入一个头节点
public void addFirstNode(int data) {
Node node = new Node(data);
node.next = first;
first = node;
}
// 删除一个头结点,并返回头结点
public Node deleteFirstNode() {
Node tempNode = first;
first = tempNode.next;
return tempNode;
}
// 在任意位置插入节点 在index的后面插入
public void add(int index, int data) {
Node node = new Node(data);
Node current = first;
Node previous = first;
while (pos != index) {
previous = current;
current = current.next;
pos++;
}
node.next = current;
previous.next = node;
pos = 0;
}
// 删除任意位置的节点
public Node deleteByPos(int index) {
Node current = first;
Node previous = first;
while (pos != index) {
pos++;
previous = current;
current = current.next;
}
if (current == first) {
first = first.next;
} else {
pos = 0;
previous.next = current.next;
}
return current;
}
// 根据节点的data删除节点(仅仅删除第一个)
public Node deleteByData(int data) {
Node current = first;
Node previous = first; // 记住上一个节点
while (current.data != data) {
if (current.next == null) {
return null;
}
previous = current;
current = current.next;
}
if (current == first) {
first = first.next;
} else {
previous.next = current.next;
}
return current;
}
// 显示出所有的节点信息
public void displayAllNodes() {
Node current = first;
while (current != null) {
current.display();
current = current.next;
}
System.out.println();
}
// 根据位置查找节点信息
public Node findByPos(int index) {
Node current = first;
if (pos != index) {
current = current.next;
pos++;
}
return current;
}
// 根据数据查找节点信息
public Node findByData(int data) {
Node current = first;
while (current.data != data) {
if (current.next == null)
return null;
current = current.next;
}
return current;
}
}

Finally we can do related tests through the test class:

public class TestLinkList {
public static void main(String[] args) {
LinkList linkList = new LinkList();
linkList.addFirstNode(20);
linkList.addFirstNode(21);
linkList.addFirstNode(19);
//print19,21,20
linkList.add(1, 22); //print19,22,21,20
linkList.add(2, 23); //print19,22,23,21,20
linkList.add(3, 99); //print19,22,23,99,21,20
//调用此方法会print 19,22,23,99,21,20
linkList.displayAllNodes();
}
}

At this point, for the singly linked list The operation notes are here.

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