Remove Linked List Elements LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Remove Linked List Elements in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

Use “Ctrl+F” To Find Any Questions Answer. & For Mobile User, You Just Need To Click On Three dots In Your Browser & You Will Get A “Find” Option There. Use These Option to Get Any Random Questions Answer.

About LeetCode

LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either EasyMedium, or Hard.

LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:

  • Mathematics/Basic Logical Based Questions
  • Arrays
  • Strings
  • Hash Table
  • Dynamic Programming
  • Stack & Queue
  • Trees & Graphs
  • Greedy Algorithms
  • Breadth-First Search
  • Depth-First Search
  • Sorting & Searching
  • BST (Binary Search Tree)
  • Database
  • Linked List
  • Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the ProblemRemove Linked List Elements– LeetCode Problem

Remove Linked List Elements– LeetCode Problem

Problem:

Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.

Example 1:

removelinked list
Input: head = [1,2,6,3,4,5,6], val = 6
Output: [1,2,3,4,5]

Example 2:

Input: head = [], val = 1
Output: []

Example 3:

Input: head = [7,7,7,7], val = 7
Output: []

Constraints:

  • The number of nodes in the list is in the range [0, 104].
  • 1 <= Node.val <= 50
  • 0 <= val <= 50
Remove Linked List Elements– LeetCode Solutions
Remove Linked List Elements Solution in C++:
class Solution {
 public:
  ListNode* removeElements(ListNode* head, int val) {
    ListNode dummy(0, head);
    ListNode* prev = &dummy;

    for (; head; head = head->next)
      if (head->val != val) {
        prev->next = head;
        prev = prev->next;
      }
    prev->next = nullptr;  // in case the last val == val

    return dummy.next;
  }
};

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Remove Linked List Elements Solution in Java:
class Solution {
  public ListNode removeElements(ListNode head, int val) {
    ListNode dummy = new ListNode(0, head);
    ListNode prev = dummy;

    for (; head != null; head = head.next)
      if (head.val != val) {
        prev.next = head;
        prev = prev.next;
      }
    prev.next = null; // in case the last val == val

    return dummy.next;
  }
}
Remove Linked List Elements Solution in Python:
class Solution:
  def removeElements(self, head: ListNode, val: int) -> ListNode:
    dummy = ListNode(0, head)
    prev = dummy

    while head:
      if head.val != val:
        prev.next = head
        prev = prev.next
      head = head.next
    prev.next = None

    return dummy.next
  • Time: O(n)
  • Space: O(1)

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