Java反转单链表(code)

主要是面试中可能会经常碰上该类似操作，尤其是稍大点公司，面试官可能并不在乎你能不能搞定该题，但是这类型题目最是能体现程序员的思维状态 ---一个迷糊头脑的程序员 怎能立志改变这个世界

/**
 * @author luochengcheng
 * 定义一个单链表
 */
class Node {
	//变量
	private int record;
	//指向下一个对象
	private Node nextNode;

	public Node(int record) {
		super();
		this.record = record;
	}
	public int getRecord() {
		return record;
	}
	public void setRecord(int record) {
		this.record = record;
	}
	public Node getNextNode() {
		return nextNode;
	}
	public void setNextNode(Node nextNode) {
		this.nextNode = nextNode;
	}
}

/**
 * @author luochengcheng
 *	两种方式实现单链表的反转(递归、普通)
 *	新手强烈建议旁边拿着纸和笔跟着代码画图(便于理解)
 */
public class ReverseSingleList {
	/**
	 * 递归，在反转当前节点之前先反转后续节点
	 */
	public static Node reverse(Node head) {
		if (null == head || null == head.getNextNode()) {
			return head;
		}
		Node reversedHead = reverse(head.getNextNode());
		head.getNextNode().setNextNode(head);
		head.setNextNode(null);
		return reversedHead;
	}

	/**
	 * 遍历，将当前节点的下一个节点缓存后更改当前节点指针
	 *
	 */
	public static Node reverse2(Node head) {
		if (null == head) {
			return head;
		}
		Node pre = head;
		Node cur = head.getNextNode();
		Node next;
		while (null != cur) {
			next = cur.getNextNode();
			cur.setNextNode(pre);
			pre = cur;
			cur = next;
		}
		//将原链表的头节点的下一个节点置为null，再将反转后的头节点赋给head
		head.setNextNode(null);
		head = pre;

		return head;
	}

	public static void main(String[] args) {
		Node head = new Node(0);
		Node tmp = null;
		Node cur = null;
		// 构造一个长度为10的链表，保存头节点对象head
		for (int i = 1; i < 10; i++) {
			tmp = new Node(i);
			if (1 == i) {
				head.setNextNode(tmp);
			} else {
				cur.setNextNode(tmp);
			}
			cur = tmp;
		}
		//打印反转前的链表
		Node h = head;
		while (null != h) {
			System.out.print(h.getRecord() + " ");
			h = h.getNextNode();
		}
		//调用反转方法
		head = reverse2(head);
		System.out.println("\n**************************");
		//打印反转后的结果
		while (null != head) {
			System.out.print(head.getRecord() + " ");
			head = head.getNextNode();
		}
	}
}