classSolution{publicNodecopyRandomList(Node head){if(head==null)returnnull;HashMap<Node,Node>map=newHashMap<>();//借助hashMapNode newHead=newNode(0);//虚拟头节点Node cur=newHead;//指针指向当前节点Node tmp;while(head!=null){if(map.containsKey(head)){//查询原节点是否已存在于maptmp=map....
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...
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://
Node*copyRandomList(Node* head){if(!head)returnhead;//copy new node and changing new node next pointer to old node next pointer and old node next pointer to new nodeNode *newHead =NULL, *l1, *l2;for(l1 = head; l1 !=NULL; l1 = l1->next->next){ l2 = new Node(l1->val, l1...
Return a deep copy of the list. 思路: 先拷贝原链表值,然后找到原链表每个结点的随机指针在链表中的位置,找到拷贝链表的随机指针的指向。 算法: 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* thisNewNode = head->next; RandomListNode* nextNewNode = head->next->next->next; thisNewNode->next = nextNewNode; //link the old list head->next = nextOldNode; head = nextOldNode; //skip one every time, head->next must not be null ...
public RandomListNode copyRandomList2(RandomListNode head) { if(head==null) return null; RandomListNode current = head, copyHead = null, copyCurrent = null; while(current!=null){ RandomListNode temp = new RandomListNode(current.label); ...
[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
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,删除其中任何一个变量,另一个变量中就没有了...
LeetCode 138. Copy List with Random Pointer 又是copy 指针的题目。 这个和上一道题目有个坑点,函数中的参数要加&地址符。 AI检测代码解析 class Solution { public: RandomListNode* ans; map<int,RandomListNode*> m; RandomListNode *copyRandomList(RandomListNode *head) { ...