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. 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. 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…

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…

原题地址 非常巧妙的方法,不需要用map,只需要O(1)的额外存储空间,分为3步: 1. 先复制链表,但是这个复制比较特殊,每个新复制的节点添加在原节点的后面,相当于"加塞"2. 根据原节点的 ramdon 指针构造新节点的 random 指针3. 恢复原链表结构,同时得到新复制链表 时间复杂度:O(n) 注意random有可能是NULL 代码: RandomListNode *copyRandomList(RandomListNode *head) { RandomListNode…

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…

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…

Copy List with Random Pointer 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/copy-list-with-random-pointer/description/ Description A linked list is given such that each node contains an additional random pointer which could point to any node in the…

题目: 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. 提示: 此题有两种方法,一种是按照原链表next的顺序依次创建节点,并处理好新链表的next指针,同时把原节点与新节点的对应关系保存到一个hash_map中,然后第…

作者: 负雪明烛 id: fuxuemingzhu 个人公众号:负雪明烛 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode-cn.com/problems/copy-list-with-random-pointer/ 题目描述 给你一个长度为 n 的链表,每个节点包含一个额外增加的随机指针 random ,该指针可以指向链表中的任何节点或空节点. 构造这个链表的 深拷贝. 深拷贝应该正好由 n 个 全…

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. Approach #1: Java. /** * Definition for singly-linked list with a random pointer. * cl…

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. 思路: 做过,先复制一遍指针,再复制random位置,再拆分两个链表. #include <iostream> #include <vector> #incl…

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. /** * Definition for singly-linked list with a random pointer. * struct RandomListNode…

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. 剑指offer里面的一道题,具体思路看不懂请查阅剑指offer /** * Definition for singly-linked list with a 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.   struct RandomListNode { int label; RandomListNode *next, *random; RandomListNode(int…

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. 思路:第一遍正常复制链表,同一时候用哈希表保存链表中原始节点和新节点的相应关系,第二遍遍历链表的时候,再复制随机域. 这是一种典型的空间换时间的做法,n个节点,须要大小为O…

题目: 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. 代码:Runtime: 215 ms # Definition for singly-linked list with a random pointer. # cl…

给出一个链表,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.返回一个深拷贝的链表. 详见:https://leetcode.com/problems/copy-list-with-random-pointer/description/ Java实现: /** * Definition for singly-linked list with a random pointer. * class RandomListNode { * int label; * Random…

职务地址:https://oj.leetcode.com/problems/copy-list-with-random-pointer/ 题意:对一个有回路的链表的深复制 解题:这道题我AC了之后才发现理解错了题意,然后就參考了以下这篇文章. 假设想看这道题的解题思路的话,请移步http://www.2cto.com/kf/201310/253477.html 拓展:假如题目给的节点不是链表的头节点.而是链表中的随意一个节点,在保证从给的点能遍历所有节点的情况下,深复制这个链表. 那么该怎么做呢…

深拷贝一个链表,不同的是这个链表有个额外的随机指针.参考:http://blog.csdn.net/ljiabin/article/details/39054999 做法非常的巧妙,分成三步,一是新建结点,并放在旧结点之后:二是修改新结点的random指针:三是将新旧链表断开. RandomListNode *randomList(RandomListNode *head) { //复制每个结点,并将新结点放在旧结点之后 for (RandomListNode *cur = head; cur…

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. 题目意思: 深拷贝一个给定的带random指针的链表,这个random指针可能会指向其他任意一个节点或者是为nu…

题目: 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. 题解: 如果要copy一个带有random pointer的list,主要的问题就是有可能这个random指向的位置还没有被copy到,所以解决方法都是多次扫描li…

类同:剑指 Offer 题目汇总索引第26题 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. 说明:分三步, 1. 每个节点复制其本身并链接在后面. 2, 复制…

LintCode - Copy List with Random Pointer LintCode - Copy List with Random Pointer Web Link Description Code - C Tips Web Link http://www.lintcode.com/en/problem/copy-list-with-random-pointer/ Description A linked list is given such that each node con…

题目 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. 分析 实现一个链表的深拷贝,返回拷贝后的新链表. 若是普通链表,逐个拷贝原始链表节点并连接即可,只需O(n)的时间复杂度:但是此题特殊的是,每个节点都有一个rando…

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. 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. 示例: 输入: {"$id":"…

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…

原题地址:https://oj.leetcode.com/problems/copy-list-with-random-pointer/ 题意: 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:这题主要是需要深…

给大家推荐一道leetcode上的面试题,这道题的详细解说在<剑指offer>的P149页有思路解说.假设你手头有这本书.建议翻阅. 题目链接 here 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. RandomLi…