utm medium list

A linked list is given such this each node contains an additional random pointer which could point to all node in the list or null.Return a deep copy of the list.Ideas:First, according to the normal order, to get a copy of the linked list, while using a hashmap to establish the corresponding relationship between the old and new nodes, and then add the random poin

Given a singly linked list l:l0→l1→ ... →ln-1→lnReorder It to:l0→ln→l1→ln-1→l2→ln-2→ ...ExampleGiven 1->2->3->4->null , reorder it to 1->4->2->3->null .ChallengeCan do this in-place without altering the nodes ' values?/*** Definition for ListNode. * public class ListNode {* int val; * ListNode Next; * ListNode (int val) {* This.val = val; * This.next = null; * } * } */ Public classSolution {/** * @paramhead:the head of linked

Given a linked list and both values V1 and v2. Swap the nodes in the linked list with values V1 and v2. It ' s guaranteed there is no duplicate values in the linked list. IF v1 or V2 does not exist in the given linked list, does nothing.NoticeYou should swaps the nodes with values V1 and v2. Do not directly swap the va

Sort a linked list in O(n log n) time using constant space complexity.ExampleGiven 1-3->2->null, sort it to 1->2->3->null./*** Definition for ListNode. * public class ListNode {* int val; * ListNode Next; * ListNode (int val) {* This.val = val; * This.next = null; * } * } */ Public classSolution {/** * @paramhead:the head of linked list. * @return: You should return the head of the sorted linked

Given a sorted linked list, delete all nodes that has duplicate numbers, leaving only distinct numbers from the Original list.ExampleGiven 1->2->3->3->4->4->5 , return 1->2->5 .Given 1->1->1->2->3 , return 2->3 ./*** Definition for ListNode * public class ListNode {* int val; * ListNode Next; * ListNode (int x) {* val = x; * next = NULL; * } * } */ Public classSolution {/** * @paramListNode Head is the head of the linked