输入一个复杂链表(每个节点中有节点值,以及两个指针,一个指向下一个节点,另一个特殊指针指向任意一个节点),返回结果为复制后复杂链表的head.(注意,输出结果中请不要返回参数中的节点引用,否则判题程序会直接返回空) 第一种方法:使用map存放原节点和其复制节点.然后再给复制节点的next/random指针初始化.初始化时,原节点p,p的复制节点的next=p.next的复制节点 见代码: public RandomListNode copyRandomList(RandomListNode he…

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…

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…

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…

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…

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…

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…

/** * Definition for singly-linked list with a random pointer. * struct RandomListNode { * int label; * RandomListNode *next, *random; * RandomListNode(int x) : label(x), next(NULL), random(NULL) {} * }; */ class Solution {//为了可以高速定位某个节点,採用确定性映射的方式,将…

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. 这题用map做其实比较简单,但是一开始没想明白越想越乱,最后看了下别人的实现,思路还是很清晰的,代码如下所示: /** * Definition for singly-li…