转载自：http://poly.iteye.com/blog/1748272
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();}}
}
)
