标签：null 链表 pre addition node could tno nal tco
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.
解题思路：分三步走，1 在每个节点后面复制该节点；2 复制随机指针 cur->next->random = cur->random->next; 3将两个链表分离。
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
标签：null 链表 pre addition node could tno nal tco
原文地址：https://www.cnblogs.com/nobodywang/p/10353641.html
