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:

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Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]

Example 2:

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Input: head = [2,1], x = 2
Output: [1,2]

Constraints:

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

Solution

Use two list node and two tail node for smaller nodes and bigger nodes. Every time add the node to new linked lists.

Time complexity: o ( n ) o(n) o(n)

Space complexity: o ( 1 ) o(1) o(1)

Code

python3 复制代码

# 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]:
        header_smaller, header2 = ListNode(-1), ListNode(-1)
        tail1, tail2 = header_smaller, header2
        p = head
        while p:
            if p.val < x:
                tail1.next = p
                tail1 = tail1.next
            else:
                tail2.next = p
                tail2 = tail2.next
            p = p.next
        tail1.next = header2.next
        tail2.next = None
        return header_smaller.next