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Problem solution · Python

Reorder List

Reorder List: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
41 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Reorder List, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 41 lines of Python from the credited upstream file 143.py.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeReorder List · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def reorderList(self, head: ListNode) -> None:    def findMid(head: ListNode):      prev = None      slow = head      fast = head       while fast and fast.next:        prev = slow        slow = slow.next        fast = fast.next.next      prev.next = None       return slow     def reverse(head: ListNode) -> ListNode:      prev = None      curr = head       while curr:        next = curr.next        curr.next = prev        prev = curr        curr = next       return prev     def merge(l1: ListNode, l2: ListNode) -> None:      while l2:        next = l1.next        l1.next = l2        l1 = l2        l2 = next     if not head or not head.next:      return     mid = findMid(head)    reversed = reverse(mid)    merge(head, reversed)

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