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[leetCode] Copy List with Random Pointer

时间:2014-11-03 20:46:00      阅读:290      评论:0      收藏:0      [点我收藏+]

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.

 

思路:首先将新节点连接到所复制的节点后面,再这样的链表上处理random指针。

   时间复杂度O(n),空间复杂度O(1)

 1 /**
 2  * Definition for singly-linked list with a random pointer.
 3  * struct RandomListNode {
 4  *     int label;
 5  *     RandomListNode *next, *random;
 6  *     RandomListNode(int x) : label(x), next(NULL), random(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     RandomListNode *copyRandomList(RandomListNode *head) {
12         if (head == NULL) return head;
13         
14         
15         for (RandomListNode *curr = head; curr != NULL;) {
16             RandomListNode *pter = new RandomListNode(curr->label);
17             pter->next = curr->next;
18             curr->next = pter;
19             curr = pter->next;
20         }
21         
22         //链接rando指针
23         for(RandomListNode *curr = head; curr != NULL;) {
24             if (curr->random != NULL) {
25                 curr->next->random = curr->random->next;
26             }
27             curr = curr->next->next;
28         }
29         
30         //分拆链表
31         RandomListNode *new_head = head->next;
32         for (RandomListNode *curr = head, **new_curr = &new_head; curr != NULL;) {
33             *new_curr = curr->next;
34             curr->next = curr->next->next;
35             
36             curr = curr->next;
37             
38             new_curr = &((*new_curr)->next);
39         }
40         
41         return new_head;
42     }
43 };

 

[leetCode] Copy List with Random Pointer

原文:http://www.cnblogs.com/vincently/p/4071991.html

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