java 实现双向链表实例详解

双向链表是一个基本的数据结构，在Java中LinkedList已经实现了这种结构，不过作为开发者，也要拥有自己显示这种结构的能力。话不多说，上代码：

首先是链表的节点类：

/** * 链表节点 * @author Administrator * * @param <T> */ public class ChainNode<T> { private T data; //对象编号 private int dataNo; public ChainNode<T> nextChainNode; public ChainNode<T> preChainNode; public ChainNode(T data, ChainNode<T> nextChainNode, ChainNode<T> preChainNode) { this.data = data; this.nextChainNode = nextChainNode; this.preChainNode = preChainNode; } public ChainNode(T data) { this.data = data; } public int getDataNo() { return dataNo; } public void setDataNo(int dataNo) { this.dataNo = dataNo; } public void setData(T data) { this.data = data; } public T getData() { return data; } }

然后是链表：

<pre name="code" class="java">/** * 链表实现类 * @author Administrator * * @param <T> */ public class Chain<T> { //头节点 private ChainNode<T> headNode; //末尾节点指针 private ChainNode<T> lastNode; private int size; //创建链表就创建头节点 public Chain() { this.headNode = new ChainNode<T>(null); this.lastNode = headNode; } //增加一个节点 public void addNode(T data) { ChainNode<T> node = new ChainNode<T>(data); if(lastNode != null){ lastNode.nextChainNode = node; node.preChainNode = node; node.setDataNo(size); lastNode = node; size++; } } //删除指定索引的节点 public void deleteNode(int dataNo) throws Exception { if(getSize() == 0){ throw new Exception("chain is empty"); } for (ChainNode<T> node = headNode.nextChainNode;node != null;node = node.nextChainNode) { if(node.getDataNo() == dataNo){ node.preChainNode.nextChainNode = node.nextChainNode; if(node.nextChainNode != null){ node.nextChainNode.preChainNode = node.preChainNode; } node.nextChainNode = null; node.preChainNode = null; size--; //重新设置节点的编号 for (ChainNode<T> chainNode = node.nextChainNode;chainNode != null;chainNode = chainNode.nextChainNode) { chainNode.setDataNo(chainNode.getDataNo()-1); } return; } } throw new Exception("the corresponding data that can not be found"); } //获取指定索引的节点 public T get(int dataNo) throws Exception { if(getSize() == 0){ throw new Exception("chain is empty"); } for (ChainNode<T> node = headNode.nextChainNode;node != null;node = node.nextChainNode) { if(node.getDataNo() == dataNo){ return node.getData(); } } throw new Exception("the corresponding data that can not be found"); } //修改对应索引的节点 public void set(int dataNo,T data) throws Exception { if(getSize() == 0){ throw new Exception("chain is empty"); } for (ChainNode<T> node = headNode.nextChainNode;node != null;node = node.nextChainNode) { if(node.getDataNo() == dataNo){ node.setData(data); return; } } throw new Exception("the data that is to be modified can not be found"); } //是否包含对应数据的节点 public boolean isContains(T data) throws Exception { if(getSize() == 0){ throw new Exception("chain is empty"); } for (ChainNode<T> chainNode = headNode.nextChainNode;chainNode != null;chainNode = chainNode.nextChainNode) { if(chainNode.getData() == data){ return true; } } return false; } //获取节点数量（不含头节点） public int getSize() { return size; } }

测试一下：

public class ChainTest { public static void main(String[] args) throws Exception{ String oneString = "one"; String twoString = "two"; String threeString = "three"; String fourString = "four"; Chain<String> chain = new Chain<String>(); chain.addNode(oneString); chain.addNode(twoString); chain.addNode(threeString); chain.addNode(fourString); for (int i = 0; i < chain.getSize(); i++) { String string2 = chain.get(i); System.out.println(string2); } try { String string = chain.get(2); System.out.println("the data of the second node is"+string); System.out.println(chain.isContains(fourString)); chain.set(1, "haha"); System.out.println("modify chain"); for (int i = 0; i < chain.getSize(); i++) { String string2 = chain.get(i); System.out.println(string2); } chain.deleteNode(3); System.out.println("delete one node"); for (int i = 0; i < chain.getSize(); i++) { String string2 = chain.get(i); System.out.println(string2); } } catch (Exception e) { // TODO Auto-generated catch block e.printStackTrace(); } } }

结果：

onetwothreefourthe data of the second node isthreetruemodify chainonehahathreefourdelete one nodeonehahathree

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