Partition List 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 Partition List 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.

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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. 

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Link for the ProblemPartition List– LeetCode Problem

Partition List– LeetCode Problem


Given the head of a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

Example 1:

Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]

Example 2:

Input: head = [2,1], x = 2
Output: [1,2]


  • The number of nodes in the list is in the range [0, 200].
  • -100 <= Node.val <= 100
  • -200 <= x <= 200
Partition List– LeetCode Solutions
Partition List Solution in C++:
class Solution {
  ListNode* partition(ListNode* head, int x) {
    ListNode beforeHead(0);
    ListNode afterHead(0);
    ListNode* before = &beforeHead;
    ListNode* after = &afterHead;

    for (; head; head = head->next)
      if (head->val < x) {
        before->next = head;
        before = head;
      } else {
        after->next = head;
        after = head;

    after->next = nullptr;
    before->next =;

Partition List Solution in Java:
class Solution {
  public ListNode partition(ListNode head, int x) {
    ListNode beforeHead = new ListNode(0);
    ListNode afterHead = new ListNode(0);
    ListNode before = beforeHead;
    ListNode after = afterHead;

    for (; head != null; head =
      if (head.val < x) { = head;
        before = head;
      } else { = head;
        after = head;
      } = null; =;

Partition List Solution in Python:
class Solution:
  def partition(self, head: ListNode, x: int) -> ListNode:
    beforeHead = ListNode(0)
    afterHead = ListNode(0)
    before = beforeHead
    after = afterHead

    while head:
      if head.val < x: = head
        before = head
      else: = head
        after = head
      head = = None =


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