力扣算法——138CopyListWithRandomPointer【M】

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.

 

Example 1:

Input:
{"$id":"1","next":{"$id":"2","next":null,"random":{"$ref":"2"},"val":2},"random":{"$ref":"2"},"val":1}

Explanation:
Node 1's value is 1, both of its next and random pointer points to Node 2.
Node 2's value is 2, its next pointer points to null and its random pointer points to itself.

 

Note:

1. You must return the copy of the given head as a reference to the cloned list.

Solution：

　　　方法一：
　　　　　使用哈希表，将链表值与节点地址对应，先复制一条主链，并用哈希表记录新链的节点值与节点地址的对应

　　　　　而后遍历原链表，经过原链表随机指针的值，和哈希表，来找到新链所对应的随机指针

　　　　缺点：须要额外的空间，而且节点值不能相同，不然哈希表失效

　　方法二：

　　　　将新链直接复制在主链后面，那么当复制随机指针时，新链的随机指针值就在旧链的随机值后面

　　　　而后将新链从主链中脱离出来

　　

 1 class Solution {
 2 public:
 3     RandomListNode *copyRandomList(RandomListNode *head) {
 4         RandomListNode* p = head;
 5         while (p != nullptr)
 6         {
 7             RandomListNode* q = new RandomListNode(p->label);
 8             q->next = p->next;
 9             p->next = q;
10             p = p->next->next;
11         }
12         p = head;
13         while (p != nullptr)
14         {
15             if (p->random != nullptr)
16                 p->next->random = p->random->next;
17             p = p->next->next;
18         }
19         p = head;
20         RandomListNode *resHead = new RandomListNode(-1), *q;
21         q = resHead;
22         while (p != nullptr)
23         {
24             q->next = p->next;
25             q = q->next;
26             p->next = p->next->next;
27             p = p->next;
28         }
29         return resHead->next;
30     }
31 };