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;
	}
};