M1: 复制带随机指针的链表

一个链表问题：复制带随机指针的链表

题目:有一个链表L,其每个节点有2个指针，一个指针next指向链表的下个节点，另一个random随机指向链表中的任一个节点，可能是自己或者为空，写一个程序，要求复制这个链表的结构并分析其复杂性

解决方法一：

O(n)的复杂度,扫面两边即可。

图【1】
图【1】是需要复制的链表

图【2】

如图【2】所示，ABCD是原来的链表，A’B’C’D’是复制的链表，第一遍扫描顺序复制next指针，把ABCD的next分别指向A’B’C’D’，将A’的next指针指向B，B’的next指针指向C，依次类推

复制random指针： A’->random=A->random->next

恢复:A->next=A’->next;A’->next=A’->next->next;

package careercup;import java.util.ArrayList;import java.util.Arrays;import java.util.HashMap;public class RandomList {public static RandomList single = new RandomList();public RandomList() {}public static void main(String[] args) {int[] arr = new int[]{1,3,5,7};RandomNode head = RandomNode.createList(arr); head.randomNext = (RandomNode)head.next.next;((RandomNode)head.next).randomNext = (RandomNode)head.next.next.next;RandomNode rn = single.getList(head);printList(rn);//printList(head);}private static void printList(RandomNode rn) {while(rn!=null){System.out.print(rn);System.out.print(", ");rn = (RandomNode)rn.next;}}public RandomNode getList(RandomNode rn) {RandomNode head = rn;RandomNode p=null;while(rn!=null){p = new RandomNode(rn);p.next = rn.next;rn.next = p;rn = (RandomNode)p.next;}p = (RandomNode)head.next;while(p!=null){p.randomNext = p.randomNext!=null ? (RandomNode)p.randomNext.next : null;if(p.next!=null) p = (RandomNode)p.next.next;else break;}rn = head;head = (RandomNode)head.next;p = head;rn.next = p.next;while(rn.next!=null){p.next = (RandomNode)p.next.next;p = (RandomNode)p.next;rn.next = p.next;}return head;}}class RandomNode extends Node{RandomNode randomNext;RandomNode(int val) {super(val);randomNext = null;}RandomNode(RandomNode rn){super(rn.val);this.next = rn.next;this.randomNext = rn.randomNext;}public static RandomNode createList(int[] arr) {RandomNode head = new RandomNode(arr[0]);RandomNode p = head;for(int i=1; i<arr.length;i++){p.next = new RandomNode(arr[i]);p = (RandomNode)p.next;}return head;}public String toString(){return "[" + val + "(" +  randomNext + ")]"; }}