LeetCode Top Interview 150
86. 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
Solution
from typing import Optional
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]:
n_head = ListNode(0)
n_tail = ListNode(0)
ptr = n_tail
temp = n_head
while head is not None:
n_next = head.next
if head.val < x:
n_head.next = head
n_head = n_head.next
else:
n_tail.next = head
n_tail = n_tail.next
head = n_next
n_tail.next = None
n_head.next = ptr.next
return temp.next