【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. 给出一个链表,每个节点包含一个额外增加的随机指针可以指向链表中的任何节点或空的节点。 返回一个深拷贝的链表。 【...
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...
【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. 给出一个链表,每个节点包含一个额外增加的随机指针可以指向链表中的任何节点或空的节点。 返回一个深拷贝的链表。 【...
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. 给出一个链表,每个节点包含一个额外增加的随机指针可以指向链表中的任何节点或空的节点。 返回一个深拷贝的链表。 【题目链接】 ww...
public class Solution { public RandomListNode copyRandomList(RandomListNode head) { if(head==null) return null; RandomListNode n1 = head; RandomListNode n2; // 生成新节点并接在旧节点后面 while(n1!=null){ n2 = new RandomListNode(n1.label); ...
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: ...
Solution: 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...
Finally, the two tabs—tab1 and tab2—have mirrored local history, as they share the same list object under the hood. Ultimately, how you implement shallow and deep copying in custom classes is entirely up to you. Additionally, if you’re using Python 3.13 or later, then you might also...
map.get(q).random = map.get(q.random); q = q.next; } return map.get(head); } } 方法二: 就是把链表弄成 一个原始节点一个copy节点,然后再循环一遍,把copy节点摘出来。 class Solution { public Node copyRandomList(Node head) {
Copy List with Random Pointer 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. 首先搞清楚什么是浅拷贝和深拷贝: 浅拷贝,把一个变量中的数据地址直接给另一个变量,删除其中任何一...