03 Jan 86. Partition List
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
# 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]:
h1 = l1 = ListNode(0)
h2 = l2 = ListNode(0)
while head:
if head.val < x:
l1.next =head
l1 = l1.next
else:
l2.next = head
l2 = l2.next
head = head.next
l2.next = None
l1.next = h2.next
return h1.next
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