[Solved] 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.

Question

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:

partition
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]

Constraints:

  • The number of nodes in the list is in the range [0, 200].
  • -100 <= Node.val <= 100
  • -200 <= x <= 200

Python Solution

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def partition(self, head: ListNode, x: int) -> ListNode:    
        l1 = ListNode(0)
        y = l1
        l2 = ListNode(0)
        z = l2
        head1 = head
        while head1:
            if head1.val < x:
                l1.next = ListNode(head1.val)
                l1 = l1.next
            else:
                l2.next = ListNode(head1.val)
                l2 = l2.next
            head1 = head1.next
        l1.next = z.next
        return y.next
Abhishek Sharma
Abhishek Sharma

Started my Data Science journey in my 2nd year of college and since then continuously into it because of the magical powers of ML and continuously doing projects in almost every domain of AI like ML, DL, CV, NLP.

Articles: 520

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