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A-035● Day 352026.06.10easyleetcode #206neetcode150

Reverse Linked List

#linked-list#recursion
↗ leetcode #206 · reverse-linked-list
01

Problem

· problem
P.035

Reverse Linked List

↗ leetcode #206

Given the head of a singly linked list, reverse the list, and return the reversed list.

constraints
  • · 0 ≤ number of nodes ≤ 5000
  • · -5000 ≤ Node.val ≤ 5000
// paraphrased summary — see source for full text
examples
example 1input → output
[1,2,3,4,5]
[5,4,3,2,1]
example 2input → output
[1,2]
[2,1]
example 3input → output
[]
[]
02

Pre-solve

· pre-solve
● 1list shown→● 2select→● 3reveal
  • ☐
    Can we reverse the list in-place, or can we create a new list?
  • ☐
    What should we return if the input is an empty list?
  • ☐
    Can we assume the input is always a valid singly linked list?
  • ☐
    Should we swap node values or just rearrange the pointers?
  • ☐
    Do we need to handle lists with cycles?
  • ☐
    Can we use the same approach for doubly linked lists?
check the items you would ask, then press confirm
// session-only state — refresh resets (repeatable practice)
03

Logic Structure

· logic
● 1slots shown→● 2pick per slot→● 3reveal
// pick one code line per slot to assemble the algorithm flow. no typing — just the logic skeleton.
step 1· Initialize Pointers
○
prev, curr = None, head
○
prev, curr = head, None
○
prev = None; curr = head.next
step 2· Loop While Nodes Exist
○
while curr:
○
while curr.next:
○
while prev:
step 3· Save Next Node│ nested
○
temp = curr.next
○
temp = prev
○
# (omitted)
step 4· Reverse the Link│ nested
○
curr.next = prev
○
prev.next = curr
○
curr.prev = prev
step 5· Advance prev│ nested
○
prev = curr
○
prev = temp
○
prev = None
step 6· Advance curr│ nested
○
curr = temp
○
curr = prev
○
curr = curr.next
step 7· Return New Head
○
return prev
○
return curr
○
return head
pick one option per slot
// format: slot — recursive / DP patterns use ordering / state-first formats. ADR-08 follow-up.
04

Solve · Trace

· solve
solution.py
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# Definition for singly-linked list.
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# class ListNode:
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#     def __init__(self, x):
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#         self.val = x
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#         self.next = None
6
7
8
class Solution:
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    def reverseList(self, head: ListNode) -> ListNode:
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        prev, curr = None, head
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        while curr:
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            temp = curr.next
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            curr.next = prev
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            prev = curr
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            curr = temp
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        return prev
mental dry-run cases
// walk each case in your head; expand the worked example below if stuck.
case 1
[1,2,3,4,5]
→
[5,4,3,2,1]
case 2
[1,2]
→
[2,1]
case 3
[]
→
[]
// UI does not walk-through — you do the dry-run mentally. Expand the worked example if stuck.