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:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def lessThan(self, head, x):
if not head:
return head
if head.val >= x:
return self.lessThan(head.next, x)
newhead = TreeNode(val = head.val)
newhead.next = self.lessThan(head.next, x)
return newhead
def greaterThan(self, head, x):
if not head:
return head
if head.val < x:
return self.greaterThan(head.next, x)
newhead = TreeNode(val = head.val)
newhead.next = self.greaterThan(head.next, x)
return newhead
def partition(self, head: Optional[ListNode], x: int) -> Optional[ListNode]:
left = self.lessThan(head, x)
right = self.greaterThan(head, x)
if not left:
return right
curr = left
while curr and curr.next:
curr = curr.next
curr.next = right
return left
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