Leetcode içerisindeki Linked List veri yapısı ile ilgili kolay soruları çözüyoruz.
Çözdüğümüz soruların linkleri:
1290. Convert Binary Number in a Linked List to Integer: https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/
876. Middle of the Linked List: https://leetcode.com/problems/middle-of-the-linked-list/
206. Reverse Linked List: https://leetcode.com/problems/reverse-linked-list/
21. Merge Two Sorted Lists: https://leetcode.com/problems/merge-two-sorted-lists/
Leetcode içerisinde bulunan “Odd Even Linked List” sorusunun açıklaması ve çözümü. Bu soruda size verilen bir linked list içerisinde, ilk node tek olacak şekilde önce tek index’teki node’ları, ardından çift index’teki node’lar gelecek şekilde düzenlemeniz isteniyor.
► LeetCode 328. Odd Even Linked List: https://leetcode.com/problems/odd-even-linked-list/
► Problem açıklaması:
Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.
The first node is considered odd, and the second node is even, and so on.
Note that the relative order inside both the even and odd groups should remain as it was in the input.
You must solve the problem in O(1) extra space complexity and O(n) time complexity.
Example 1:
Input: head = [1,2,3,4,5]
Output: [1,3,5,2,4]
Example 2:
Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]
Constraints:
n == number of nodes in the linked list
0 <= n <= 10^4
-10^6 <= Node.val <= 10^6
LeetCode içerisinde bulunan “Intersection of Two Linked Lists” sorusunun açıklaması ve çözümü. Bu soruda sizi verilen iki adet linked list‘in kesiştiği bir node var mı, varsa o node’u yoksa geriye null döndürmeniz isteniyor.
► LeetCode 160. Intersection of Two Linked Lists: https://leetcode.com/problems/intersection-of-two-linked-lists/
► Problem açıklaması:
Given the heads of two singly linked-lists headA and headB, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null.
For example, the following two linked lists begin to intersect at node c1:
The test cases are generated such that there are no cycles anywhere in the entire linked structure.
Note that the linked lists must retain their original structure after the function returns.
Custom Judge:
The inputs to the judge are given as follows (your program is not given these inputs):
intersectVal – The value of the node where the intersection occurs. This is 0 if there is no intersected node.
listA – The first linked list.
listB – The second linked list.
skipA – The number of nodes to skip ahead in listA (starting from the head) to get to the intersected node.
skipB – The number of nodes to skip ahead in listB (starting from the head) to get to the intersected node.
The judge will then create the linked structure based on these inputs and pass the two heads, headA and headB to your program. If you correctly return the intersected node, then your solution will be accepted.
Follow up: Could you write a solution that runs in O(n) time and use only O(1) memory?
Hackerrank’in 30 Days of Code içerisinde bulunan “Day 15: Linked List” sorusunun açıklaması ve çözümü. Bu soruda linked list veri yapısına ve içerisindeki node’larda iterasyon yapılmasına göz atıyoruz.
► Hackerrank 30 Days of Code Çözümleri – Day 15: Linked List: https://www.hackerrank.com/challenges/30-linked-list/problem
► Problem açıklaması:
Objective
Today we will work with a Linked List. Check out the Tutorial tab for learning materials and an instructional video.
A Node class is provided for you in the editor. A Node object has an integer data field, data, and a Node instance pointer, next, pointing to another node (i.e.: the next node in the list).
A Node insert function is also declared in your editor. It has two parameters: a pointer, head, pointing to the first node of a linked list, and an integer, data, that must be added to the end of the list as a new Node object.
Task
Complete the insert function in your editor so that it creates a new Node (pass data as the Node constructor argument) and inserts it at the tail of the linked list referenced by the head parameter. Once the new node is added, return the reference to the head node.
Note: The head argument is null for an empty list.
Input Format
The first line contains T, the number of elements to insert.
Each of the next T lines contains an integer to insert at the end of the list.
Output Format
Return a reference to the head node of the linked list.
Sample Input
STDIN Function
—– ——–
4 T = 4
2 first data = 2
3
4
1 fourth data = 1
Sample Output
2 3 4 1
Explanation
T = 4, so your method will insert 4 nodes into an initially empty list. First the code returns a new node that contains the data value 2 as the head of the list. Then create and insert nodes 3, 4, and a at the tail of the list.
HackerRank içerisinde bulunan “Cycle Detection” sorusunun açıklaması ve çözümü. Bu soruda size verilen bir linked list içerisinde döngüsel bir yapı olup olmadığını bulmanız isteniyor.
► HackerRank – Cycle Detection: https://www.hackerrank.com/challenges/detect-whether-a-linked-list-contains-a-cycle/problem
► Problem açıklaması:
A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0.
Example
head refers to the list of nodes 1 — 2 — 3 — NULL
The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0.
head refers to the list of nodes 1 — 2 — 3 — 1 — NULL
There is a cycle where node 3 points back to node 1, so return 1.
Function Description
Complete the has_cycle function in the editor below.
It has the following parameter:
SinglyLinkedListNode pointer head: a reference to the head of the list
Returns
int:1 if there is a cycle or 0 if there is not
Note: If the list is empty, head will be null.
Input Format
The code stub reads from stdin and passes the appropriate argument to your function. The custom test cases format will not be described for this question due to its complexity. Expand the section for the main function and review the code if you would like to figure out how to create a custom case.