链表的反转（递归实现和非递归实现）

方法一：非递归实现，使用三个指针。

Java代码如下：

package com.yiguy.test;

class Node{
	int data;
	Node next;
	
	Node(){
	}
	
	Node(int dataParam){
		this.data = dataParam;
		this.next = null;
	}
}

public class List {
	// 反转链表，返回值为：反转之后的头结点
	public static Node reverse(Node head){
		if(head == null || head.next == null){
			return head;
		} 
		// 头结点的反转单独处理
		Node previous = head;
		Node current = previous.next;
		Node next;
		if(current != null){
			next = current.next;
			current.next = previous;
			previous.next = null;
			// 处理剩余部分
			previous = current;
			current = next;
			while(current != null){
				next = current.next;
				current.next = previous;
				previous = current;
				current = next;
			}
		}
		return previous;
	}
	
	public static void main(String[] args) {

		Node head = new Node(1);
		Node node1 = new Node(2);
		Node node2 = new Node(3);
		Node node3 = new Node(4);
		Node node4 = new Node(5);
		head.next = node1;
		node1.next = node2;
		node2.next = node3;
		node3.next = node4;
		Node newHead = List.reverse(head);
		while(newHead != null){
			System.out.print(newHead.data + " ");
			newHead = newHead.next;
		}
		
	}
}

方法二：递归实现

package com.yiguy.test;

class Node {
	int data;
	Node next;

	Node() {
	}

	Node(int dataParam) {
		this.data = dataParam;
		this.next = null;
	}
}

public class List {
	// 反转链表，返回值为：反转之后的头结点
	public static Node reverse(Node head) {
		if (head == null || head.next == null) {
			return head;
		}
		Node newHead = reverse(head.next);
		head.next.next = head;
		head.next = null;
		return newHead;
	}

	public static void main(String[] args) {

		Node head = new Node(1);
		Node node1 = new Node(2);
		Node node2 = new Node(3);
		Node node3 = new Node(4);
		Node node4 = new Node(5);
		head.next = node1;
		node1.next = node2;
		node2.next = node3;
		node3.next = node4;
		Node newHead = List.reverse(head);
		while (newHead != null) {
			System.out.print(newHead.data + " ");
			newHead = newHead.next;
		}

	}
}