Linked-List

A linked list of length n is given such that each node contains an additional random pointer, which could point to any node in the list, or null. Construct a deep copy of the list. The deep copy should consist of exactly n brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next and random pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. None of the pointers in the new list should point to nodes in the original list. ...

Convert a Binary Search Tree to a sorted Circular Doubly-Linked List in place. You can think of the left and right pointers as synonymous to the predecessor and successor pointers in a doubly-linked list. For a circular doubly linked list, the predecessor of the first element is the last element, and the successor of the last element is the first element. We want to do the transformation in place. After the transformation, the left pointer of the tree node should point to its predecessor, and the right pointer should point to its successor. You should return the pointer to the smallest element of the linked list. ...

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order. Merge all the linked-lists into one sorted linked-list and return it. Example 1: Input: lists = [[1,4,5],[1,3,4],[2,6]] Output: [1,1,2,3,4,4,5,6] Explanation: The linked-lists are: [ 1->4->5, 1->3->4, 2->6 ] merging them into one sorted list: 1->1->2->3->4->4->5->6 Example 2: Input: lists = [] Output: [] Example 3: Input: lists = [[]] Output: [] Constraints: ...

For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees. A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations. Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivelent or false otherwise. ...

21. Merge Two Sorted Lists Merge two sorted linked lists and return it as a sorted list. The list should be made by splicing together the nodes of the first two lists. Example 1: Input: l1 = [1,2,4], l2 = [1,3,4] Output: [1,1,2,3,4,4] Example 2: Input: l1 = [], l2 = [] Output: [] Example 3: Input: l1 = [], l2 = [0] Output: [0] ...

1625. Lexicographically Smallest String After Applying Operations Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well. Example 1: Input: head = [1,1,2] Output: [1,2] Example 2: Input: head = [1,1,2,3,3] Output: [1,2,3] Constraints: The number of nodes in the list is in the range [0, 300]. -100 <= Node.val <= 100 The list is guaranteed to be sorted in ascending order. Solution class Solution { String smallest = null; public String findLexSmallestString(String s, int a, int b) { smallest = s; Set<String> set = new HashSet<>(); dfs(s, set, a, b); return smallest; } void dfs(String current, Set<String> set, int a, int b) { if (set.contains(current)) { return; } set.add(current); if (current.compareTo(smallest) < 0) { smallest = current; } dfs(add(current, a), set, a, b); dfs(rotate(current, b), set, a, b); } String add(String s, int a) { StringBuilder sb = new StringBuilder(s); for (int i = 1; i < s.length(); i += 2) { char c = sb.charAt(i); int value = c - '0'; value += a; value = value % 10; sb.deleteCharAt(i); sb.insert(i, value); } return sb.toString(); } String rotate(String s, int b) { int len = s.length(); return s.substring(len - b, len) + s.substring(0, len - b); } }

206. Reverse Linked List Given the head of a singly linked list, reverse the list, and return the reversed list. Example 1: Input: head = [1,2,3,4,5] Output: [5,4,3,2,1] Example 2: Input: head = [1,2] Output: [2,1] Example 3: Input: head = [] Output: [] Constraints: The number of nodes in the list is the range [0, 5000]. -5000 <= Node.val <= 5000 Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both? ...

24. Swap Nodes in Pairs Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list’s nodes (i.e., only nodes themselves may be changed.) Example 1: Input: head = [1,2,3,4] Output: [2,1,4,3] Example 2: Input: head = [] Output: [] Example 3: Input: head = [1] Output: [1] Constraints: The number of nodes in the list is in the range [0, 100]. 0 <= Node.val <= 100 Solution /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solution { public ListNode swapPairs(ListNode head) { ListNode dummy = new ListNode(0); dummy.next = head; ListNode runner = dummy; while (runner.next != null && runner.next.next != null) { ListNode r1 = runner.next; ListNode r2 = runner.next.next; runner.next = r2; r1.next = r2.next; r2.next = r1; runner = runner.next.next; } return dummy.next; } }

237. Delete Node in a Linked List Write a function to delete a node in a singly-linked list. You will not be given access to the head of the list, instead you will be given access to the node to be deleted directly. It is guaranteed that the node to be deleted is not a tail node in the list. Example 1: Input: head = [4,5,1,9], node = 5 Output: [4,1,9] Explanation: You are given the second node with value 5, the linked list should become 4 -> 1 -> 9 after calling your function. ...

19. Remove Nth Node From End of List Given the head of a linked list, remove the nth node from the end of the list and return its head. Example 1: Input: head = [1,2,3,4,5], n = 2 Output: [1,2,3,5] Example 2: Input: head = [1], n = 1 Output: [] Example 3: Input: head = [1,2], n = 1 Output: [1] Constraints: The number of nodes in the list is sz. 1 <= sz <= 30 0 <= Node.val <= 100 1 <= n <= sz Solution /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { ListNode dummy = new ListNode(0); dummy.next= head; ListNode walker = dummy; ListNode runner = dummy; for (int i = 0; i < n + 1; i++) { runner = runner.next; } while (runner != null) { walker = walker.next; runner = runner.next; } walker.next = walker.next.next; return dummy.next; } }