算法描述：

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.

解題思路：分三步走，1 在每個節點後面復制該節點；2 復制隨機指針 cur->next->random = cur->random->next; 3將兩個鏈表分離。

    RandomListNode *copyRandomList(RandomListNode *head) {
        
 
if(head == nullptr) return head;
        RandomListNode* cur = head;
        while(cur!=nullptr){
            RandomListNode* temp = new RandomListNode(cur->label);
            temp->next = cur->next;
            cur->next = temp;
            cur = cur->next->next;
        }
        RandomListNode
 
* newHead = head->next;
        cur = head;
        while(cur!=nullptr){
            if(cur->random!=nullptr)
                cur->next->random = cur->random->next;
            cur = cur->next->next;
        }
        cur = head;
        RandomListNode* ncur = newHead;
        
 
while(cur!=nullptr){
            cur->next = ncur->next;
            cur = cur->next;
            if(cur!=nullptr){
                ncur->next = cur->next;
                ncur= ncur->next;
            }
        }
        return newHead;
    }

LeetCode-138-Copy List with Random Pointer