图解: 此题过程分为三个阶段,分别是 1.负责后面一个节点,并且将这个节点插入到原来链表中  2.复制后面一个节点的random指针. 3 拆分组合链表为两部分. 第一部分代码: while(currentnode!=null){//复制节点,并且将节点插入到该节点的后面 RandomListNode clonenode=new RandomListNode(currentnode.label); clonenode.next=currentnode.next; currentnode.next…

Total Accepted: 53943 Total Submissions: 209664 Difficulty: Hard 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.  (M) Clone Graph   o(…

Each of the nodes in the linked list has another pointer pointing to a random node in the list or null. Make a deep copy of the original list. /** * class RandomListNode { * public int value; * public RandomListNode next; * public RandomListNode rand…

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…

一直直到bug-free.不能错任何一点. 思路不清晰:刷两天. 做错了,刷一天. 直到bug-free.高亮,标红. 185,OA(YAMAXUN)--- (1) findFirstDuplicate string in a list of string. import java.util.HashSet; import java.util.Set; public class Solution { public static void main(String[] args) { String[…

问题: 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.   结点的定义如下: /** * Definition for singly-linked list with a random pointer. * class…

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. 随机链表的节点数据结构 struct RandomListNode { int label; RandomListNode *next, *random; RandomLi…

463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…

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. Solution: Uses map to recorde node mapping between old linked list and new linked list…