143-reorder-list

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Description

You are given the head of a singly linked-list. The list can be represented as:

L0 → L1 → … → Ln - 1 → Ln

Reorder the list to be on the following form:

L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …

You may not modify the values in the list's nodes. Only nodes themselves may be changed.

 

Example 1:

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

Example 2:

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

 

Constraints:

• The number of nodes in the list is in the range [1, 5 * 104].
• 1 <= Node.val <= 1000

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 reorderList(self, head: Optional[ListNode]) -> None:
        """
        Do not return anything, modify head in-place instead.
        """
        def findmiddle(head):
            slow = head
            fast = head
            
            while fast and fast.next:
                slow = slow.next
                fast = fast.next.next
            return slow.next, slow
        
        first = head
        second_half_start,first_end = findmiddle(first)
        
        def reverse(head):
            cur = head
            prev = None
            while cur:
                nxt = cur.next
                cur.next = prev
                prev = cur
                cur = nxt
            return prev
        
        if first_end is None:
            return head
        first_end.next = None
        
        second_half_start = reverse(second_half_start)
        
        def merge(head1, head2):
            res = cur = ListNode(None)
            
            while head1 or head2:
                if head1:
                    res.next = head1
                    head1 = head1.next
                    res = res.next
                if head2:
                    res.next = head2
                    head2 = head2.next
                    res = res.next
                
            while head1:
                res.next = head1
                head1 = head1.next
                res = res.next
            while head2:
                res.next = head2
                head2 = head2.next
                res = res.next
            
            return res.next
            
            
        head = merge(first, second_half_start)