复杂链表的复制

声明：本篇文章是阅读了何海涛的日志后，自己用java 实现后的学习总结。

题目：有一个复杂链表，其结点除了有一个Next指针指向下一个结点外，还有一个Sibling指向链表中的任一结点或者NULL。

思路：方法一：第一遍先复制节点，复制next和val，后面再单独处理sibling，这个时间复杂度为O（n*n）

方法二：将复制出来的节点穿插在原节点中间（对应何海涛日志的第三种方法）。然后用O(n)的时间来处理sibling。

何海涛日志：http://zhedahht.blog.163.com/blog/static/254111742010819104710337/

代码：

/** * 题目：有一个复杂链表，其结点除了有一个m_pNext指针指向下一个结点外，还有一个m_pSibling指向链表中的任一结点或者NULL。 * @author hongbin.gao *  * */class ComplexNode{  //复杂链表节点int val;ComplexNode next;ComplexNode sibling;public ComplexNode(int val){this.val = val;}}public class CloneComplexNodes {public static void main(String[] args){//程序中的例子和何海涛日志中的例子相同，我只是将ABCDE换成了1、2、3、4、5而已。ComplexNode head = new ComplexNode(1);ComplexNode p1 = new ComplexNode(2);ComplexNode p2 = new ComplexNode(3);ComplexNode p3 = new ComplexNode(4);ComplexNode p4 = new ComplexNode(5);head.next = p1;p1.next = p2;p2.next = p3;p3.next = p4;p4.next = null;head.sibling = p2;p1.sibling = p4;p3.sibling = p1;p2.sibling = null;p4.sibling = null;ComplexNode cloneHead = cloneComplexNodesII(head);while(cloneHead!= null){System.out.print("val="+cloneHead.val + ",");if(cloneHead.next == null)System.out.print("next=null,");else   System.out.print("next=" + cloneHead.next.val + ",");if(cloneHead.sibling == null)System.out.println("sibling=null;");else     System.out.println("sibling="+ cloneHead.sibling.val + ";");cloneHead = cloneHead.next;}}/** * cloneComplexNodesI是按照常规的思路来编写的，先复制主题，再复制sibling，时间复杂度为O(n*n); * @param head * @return */public static ComplexNode cloneComplexNodesI(ComplexNode head){  //复制复杂链表if(head == null)return null;ComplexNode headp = head;ComplexNode root = new ComplexNode(headp.val); //开始复制第一个节点ComplexNode rootp =root;headp = headp.next;while(headp != null){  //先复制复杂链表的主题部分，和next。ComplexNode p = new ComplexNode(headp.val);rootp.next = p;rootp = rootp.next;headp = headp.next;}ComplexNode headp2 = headp = head;ComplexNode rootp2 = rootp = root;int index = 0; //和头结点的距离int num1 = 0; //表示headp现在位于哪里int num2 = 0; //表示rootp现在位于哪里while(headp!= null){index = 0; //index用来表示对应的sibling是链表的第几个节点while(headp.next!=null && headp.sibling == null){headp = headp.next;num1 ++;}headp2 =head;while(headp2 != headp.sibling){headp2 =headp2.next;index ++;}while(rootp.next!=null && num2<num1){rootp = rootp.next;rootp.sibling = null;num2 ++;}rootp2 = root;while(index >0){rootp2 = rootp2.next;index--;}rootp.sibling = rootp2;headp = headp.next;num1++;rootp = rootp.next;num2++;}return root;}/** * 对应何海涛的方法三，将新复制的节点连接在原节点后面，最后再将链表拆开为两个链表。 * @param head * @return */public static ComplexNode cloneComplexNodesII(ComplexNode head){  //复制复杂链表if(head == null)return null;//先复制节点ComplexNode headp = head;while(headp != null){ComplexNode p = new ComplexNode(headp.val);p.next = headp.next;headp.next = p;headp = p.next;}//处理siblingheadp = head;ComplexNode rootp = null; //拆分过程中处于链表的尾部，进行操作。rootp  = headp.next;while(headp !=null){if(headp.sibling != null)rootp.sibling = headp.sibling.next;elserootp.sibling = null;headp = rootp.next;if(headp !=null)   rootp = headp.next;}//拆链表ComplexNode root = null; //要返回的指向新复制的链表的头结点。rootp = null;headp = head;root = headp.next;headp.next = root.next;rootp = root;headp = headp.next;while(headp!=null){rootp.next = headp.next;rootp = rootp.next;headp.next = rootp.next;headp = headp.next;}rootp.next = null;return root;}}