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A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.
Return a deep copy of the list.
分析：
这题的关键其实是如何copy random pointer，找到一个random pointer指向的node, time complexity is O(n). 所以总的复杂度在O(n^2).
这里有一个比较巧妙的方法。在每个Node后面添加一个Node，新node的值和前一个Node一样。这样的话random pointer就很容易copy了，
1 /** 2 * Definition for singly-linked list with a random pointer. 3 * class RandomListNode { 4 * int label; 5 * RandomListNode next, random; 6 * RandomListNode(int x) { this.label = x; } 7 * }; 8 */ 9 public class Solution { 10 /** 11 * @param head: The head of linked list with a random pointer. 12 * @return: A new head of a deep copy of the list. 13 */ 14 public RandomListNode copyRandomList(RandomListNode listHead) { 15 if (listHead == null) return listHead; 16 RandomListNode current = listHead; 17 // add duplicate nodes 18 while(current != null) { 19 RandomListNode node = new RandomListNode(current.label); 20 node.next = current.next; 21 current.next = node; 22 current = current.next.next; 23 } 24 25 // set random pointer 26 current = listHead; 27 while (current != null) { 28 if (current.random != null) { 29 current.next.random = current.random.next; 30 } 31 current = current.next.next; 32 } 33 34 // restore and detach 35 RandomListNode newHead = listHead.next; 36 RandomListNode current1 = listHead; 37 RandomListNode current2 = newHead; 38 while (current1 != null) { 39 current1.next = current2.next; 40 current1 = current1.next; 41 if (current1 != null) { // very important, cannot remove it. 42 current2.next = current1.next; 43 } 44 current2 = current2.next; 45 } 46 return newHead; 47 } 48 }
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5636515.html
