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. 题意:普通的链表中给定一个随机指针,该指针可以为空,也可以指向链表中任意节点。要求对该链表深拷贝。 思路: 与普通的链表拷贝不...
下面分别为算法1和算法2代码: 1classSolution {2public:3RandomListNode *copyRandomList(RandomListNode *head)4{5//Note: The Solution object is instantiated only once and is reused by each test case.6if(head == NULL)returnNULL;7std::map<RandomListNode *,RandomListNode *>oldlistMap;8RandomListNo...
public RandomListNode copyRandomList(RandomListNode head) { // head2表示拷贝链表表头,t指向拷贝链表最后一个结点 RandomListNode p = head, head2 = null, q = null, t = head2; // ===建立链表值的拷贝链表 while (p != null) { q = new RandomListNode(p.label); if (head2 == null) { ...
}//copy random pointerRandomListNode *q =head;while(q){ RandomListNode*tmp = q->next;if(q->random){ tmp->random = q->random->next; } q=tmp->next; }//seperate listRandomListNode *dupHead = head == NULL ? NULL : head->next, *cur =head;while(cur){ RandomListNode*tmp = cur-...
Return a deep copy of the list. 栈迭代 复杂度 时间O(N) 空间 O(N) 如果不算新链表的空间则是O(1) 思路 由于随机指针有可能产生环路,我们不能直接沿着随机指针的方向一个一个复制。同时我们又不能沿着next指针直接复制,因为我们不知道随机指针所指向的新节点是哪个。这里有个技巧,我们把复制的表一对一的...
[前端]-[刷题]-leetcode 138 copy-list-with-random-pointer, 视频播放量 15、弹幕量 0、点赞数 0、投硬币枚数 0、收藏人数 0、转发人数 0, 视频作者 rhythm022, 作者简介 ,相关视频:[前端]-[刷题]-leetcode 1721 swapping-nodes-in-a-linked-list,[前端]-[刷题]-leetcod
publicNodecopyRandomList(Nodehead){if(head==null){returnnull;}Nodel1=head;Nodel2=null;//生成所有的节点,并且分别插入到原有节点的后边while(l1!=null){l2=newNode(l1.val);l2.next=l1.next;l1.next=l2;l1=l1.next.next;}//更新插入节点的 randoml1=head;while(l1!=null){if(l1.random!=null)...
Node.random 为空(null)或指向链表中的节点。 节点数目不超过 1000 。 思路: 方法1: Hashmap字典映射 方法2: 新节点连接到原始节点(node1->node1'->node2->node2') 新节点random指针赋值 拆分新旧链表 代码: class Solution: def copyRandomList(self, head: Node) -> Node: ...
publicRandomListNodecopyRandomList(RandomListNodehead){if(head==null)returnnull;RandomListNodeiter=head;while(iter!=null){RandomListNodecopy=newRandomListNode(iter.label);copy.next=iter.next;iter.next=copy;iter=iter.next.next;}iter=head;while(iter!=null){if(iter.random!=null){iter.next.random=...
Can you solve this real interview question? Copy List with Random Pointer - A linked list of length n is given such that each node contains an additional random pointer, which could point to any node in the list, or null. Construct a deep copy [https://