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. Example 1: Input: {"$id":"1","next":{"$id":"2&…

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. 这道题的难点在于如何处理随机指针,由于每一个节点都有一个随机指针,这个指针可以为空,也可以指向链表的任意一个节点,如果在生成一个新节点给其随机指针赋值时,都去遍历原链表的话…

public RandomListNode copyRandomList(RandomListNode head) { /* 深复制,就是不能只是复制原链表变量,而是做一个和原来链表一模一样的新链表, 每一个节点都是新建的,而不是指向就节点 这个题的难点在于:随机节点. 随机节点有可能指向后边还没有建立的节点,这就没法指. 方法一:一一对应地记录每个新旧节点的映射关系,在常规节点建立后,就去查哈希表,找到对应 新节点的旧节点的random,就是新节点的random */ if (head==nu…

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. 这道链表的深度拷贝题的难点就在于如何处理随机指针的问题,由于每一个节点都有一个随机指针,这个指针可以为空,也可以指向链表的任意一个节点,如果我们在每生成一个新节点给其随机指…

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. 深度复制: 仅简单的遍历一遍链表时,没法复制random pointer属性.所以有点懵,大神的做法如下,加入个人理解. 思路:对链表进行三次遍历. 第一次遍历复制每一个结…

题目描述: 中文: 给定一个链表,每个节点包含一个额外增加的随机指针,该指针可以指向链表中的任何节点或空节点. 要求返回这个链表的深拷贝. 示例: 输入:{"$id":"1","next":{"$id":"2","next":null,"random":{"$ref":"2"},"val":2},"…

133. Clone Graph Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors. OJ's undirected graph serialization: Nodes are labeled uniquely. We use # as a separator for each node, and , as a separator for node lab…

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: 我们在Java for LeetCode 133 Clone Graph题中做过图的复制,本题和图的复制十分类似,JAVA实现如下: public Random…

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. 和第133题差不多,都是图的复制,区别在于这道题的label有可能是相同的,所以导致了map的key有可能相同,所以需要处理. 两种方法差不多.第二种更简洁. 1.在复制n…

python代码如下: # Definition for singly-linked list with a random pointer. # class RandomListNode(object): # def __init__(self, x): # self.label = x # self.next = None # self.random = None class Solution(object): def copyRandomList(self, head): ""&q…