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86. Partition List

Linked List Two Pointers

Problem - Partition List

Medium

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]

 

Constraints:

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

Solutions

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# 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: Optional[ListNode], x: int) -> Optional[ListNode]:
        left = ListNode()
        right = ListNode()
        temp_right, temp_left = right, left
        while head:
            if head.val < x:
                temp_left.next = head
                temp_left = temp_left.next
            else:
                temp_right.next = head
                temp_right = temp_right.next
            head = head.next

        temp_right.next = None
        temp_left.next = right.next
        return left.next

Submission Stats:

  • Runtime: 0 ms (100.00%)
  • Memory: 17.8 MB (65.35%)