标签：style blog http color io div 算法 new
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.
剑指offer中的原题
算法思路：
遍历原list的每一个节点，对每一个节点生成一个copy节点，插到原节点的后面。完成next的拷贝
第二遍扫描，对每一个原节点的random节点，其copy节点的random应该为原节点random.next。完成random的拷贝
第三遍扫描，将copy list取下来。
【注意】：分解原list和copy list时候，要完整的把原list给组装起来。不能破坏原list的结构。
1 public class Solution { 2 public RandomListNode copyRandomList(RandomListNode head) { 3 if(head == null) return null; 4 RandomListNode node = head; 5 while(node != null){ 6 RandomListNode tem = new RandomListNode(node.label); 7 tem.next = node.next; 8 node.next = tem; 9 node = node.next.next; 10 } 11 node = head; 12 while(node != null){ 13 RandomListNode next = node.next; 14 if(node.random != null) 15 next.random = node.random.next; 16 node = node.next.next; 17 } 18 RandomListNode hhead = new RandomListNode(0); 19 RandomListNode pointer = hhead; 20 node = head; 21 while(node != null){ 22 pointer.next = node.next; 23 pointer = node.next; 24 node.next = node.next.next; 25 node = node.next; 26 } 27 return hhead.next; 28 } 29 }
[leetcode]Copy List with Random Pointer,布布扣,bubuko.com
[leetcode]Copy List with Random Pointer
标签：style blog http color io div 算法 new
原文地址：http://www.cnblogs.com/huntfor/p/3894854.html
