86-partition-list

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Description

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

Solution(Python)

# 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]:
        before_head = before = ListNode()
        after_head = after = ListNode()

        cur = head

        while cur:
            if cur.val < x:
                before.next = cur
                before = before.next
            else:
                after.next = cur
                after = after.next
            cur = cur.next
        after.next = None
        before.next = after_head.next
        return before_head.next