Given a node from a cyclic linked list which is sorted in ascending order, write a function to insert a value into the list such that it remains a cyclic sorted list. The given node can be a reference to any single node in the list, and may not be necessarily the smallest value in the cyclic list. If there are multiple suitable places for insertion, you may choose any place to insert the new value.
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. You may assume the two numbers do not contain any leading zero, except the number 0 itself. Example Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807.
Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put. get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1. put(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 Solution: /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { if (l1 == null) return l2; if (l2 == null) return l1; ListNode dummy = new ListNode(0); ListNode temp = dummy; while (l1 !
Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up: Can you solve it without using extra space? Observation: Solution: /** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode detectCycle(ListNode head) { if (head == null) return null; ListNode slow = head; ListNode fast = head; boolean hasCycle = false; while (slow.
Given a linked list, determine if it has a cycle in it. Follow-up Can you solve it without using extra space? Solution # Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def hasCycle(self, head): """ :type head: ListNode :rtype: bool """ slow = head fast = head while fast: # fast reaches the end, no cycle was detected if fast and not fast.
Write a program to find the node at which the intersection of two singly linked lists begins. For example, the following two linked lists: A: a1 → a2 ↘ c1 → c2 → c3 ↗ B: b1 → b2 → b3 begin to intersect at node c1. Note If the two linked lists have no intersection at all, return null. The linked lists must retain their original structure after the function returns.
Reverse a linked list from position m to n. Do it in one-pass. Note: 1 ≤ m ≤ n ≤ length of list. Example: Input: 1->2->3->4->5->NULL, m = 2, n = 4 Output: 1->4->3->2->5->NULL Solution: /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode reverseBetween(ListNode head, int m, int n) { ListNode dummy = new ListNode(0); dummy.
Remove all elements from a linked list of integers that have value val. Example Input: 1->2->6->3->4->5->6, val = 6 Output: 1->2->3->4->5 Solution # Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def removeElements(self, head, val): """ :type head: ListNode :type val: int :rtype: ListNode """ pre = ListNode(0) pre.next = head pos = pre while pos.next != None: if pos.
Given a sorted linked list, delete all duplicates such that each element appear only once. Example 1 Input: 1->1->2 Output: 1->2 Example 2 Input: 1->1->2->3->3 Output: 1->2->3 Solution # Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def deleteDuplicates(self, head): """ :type head: ListNode :rtype: ListNode """ curr = head while curr != None and curr.next != None: if curr.