【转】JAVA实现两种方法反转单链表

http://www.java3z.com/cwbwebhome/article/article5/52407.html?id=4839

/** * @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(); } } }

运行: 
C:\ex>java   ReverseSingleList 
0 1 2 3 4 5 6 7 8 9 
************************** 
9 8 7 6 5 4 3 2 1 0 

posted @ 2015-10-09 21:18  小东瓜刨冰  阅读(302)  评论(0编辑  收藏  举报