Copy List with Random Pointer

题目描述：

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.
解题思路：
使用hashtable保存原始链表中的结点和新生成的链表中的结点的对应关系。

然后根据该对应关系生成新链表中各结点的random pointer

AC代码如下：

class Solution {
public:
RandomListNode *copyRandomList(RandomListNode *head) {
map<RandomListNode*, RandomListNode*> mapping;
RandomListNode* p = head;
RandomListNode* dummyHead = new RandomListNode(-1);
RandomListNode* q = dummyHead;
while (p){
q->next = new RandomListNode(p->label);
q = q->next;
mapping[p] = q;
p = p->next;
}
p = head;
while (p){
if (p->random != NULL){
mapping[p]->random = mapping[p->random];
}
p=p->next;
}
return dummyHead->next;
}
};