Given the head of a linked list, reverse the list and return the new head.
Input: [1,2,3,4,5]
Output: [5,4,3,2,1]
Input: [1,2]
Output: [2,1]
Input: [1]
Output: [1]
Constraints
Let's start our approach with a concrete understanding of how a linked list works. Unlike an array, a linked list is a recursive data structure - each node points to another node - which means we don't have random access to its members. Instead, we're given a head node with a next
property that points to the subsequent node in the list. Here's an example of a linked list with two nodes:
class Node:
def _init_(self, value=None):
self.value = value
self.next = None
n1 = Node(1)
n2 = Node(2)
n1.next = n2
If this list were to be reversed, its clear that n2
would point to n1
. But what would n1
point to? Recall that while n2
does not have a child, the node's next
property still exists - it simply points to null
. And while there is no node pointing to n1
, we can imagine that its parent is null
. So n1
would point to null
.
Caption: We can imagine a
null
node at the head and the tail of the list.
At minimum, reversing a linked list will require updating each node's next
pointer to reference its parent. Since we'll need to visit each node at least once, our solution space is limited to a linear traversal.
Let's explore how we can update each node in-place with a linear traversal. A linked list can be traversed both recursively and iteratively - in both approaches, we maintain a reference to the current node, its parent, and its child, and re-assign the next
reference for each node.
Since a linked list is an inherently recursive data structure, it makes sense to consider a post-order recursive traversal to reverse the list. Why "post-order"? The key here is to recurse on each subsequent node until the last node is reached, and then update the next
pointers as each execution pops off the call stack.
Each call to reverse_list
is passed in a reference to the current node's child, which adds a new execution frame to the call stack.
Once a null
node is reached (our base case), we begin the reassignment process. As each context is popped off the stack, we assign the current node's child's next
pointer to the current node, effectively reversing its reference. Then return the reversed list so far.
We also set the current node's next
pointer to null
- this will be overwritten once the subsequent recursive call is resolved, except for the original head which is now the tail and therefore has no next
node.
class Solution:
def reverse_list(curr_node: Node) -> Node:
if not curr_node or not curr_node.next:
return curr_node
prev = self.reverse_list(curr_node.next)
curr_node.next.next = curr_node
curr_node.next = None
return prev
1class Solution:
2 def reverse_list(curr_node: Node) -> Node:
3 if not curr_node or not curr_node.next:
4 return curr_node
5 prev = self.reverse_list(curr_node.next)
6 curr_node.next.next = curr_node
7 curr_node.next = None
8 return prev
O(n)
O(n)
Recursion requires linear space on the call stack to maintain a reference to each execution context.Reversing a linked list iteratively is more space efficient than recursion, and tends to be more easy to grasp. The task of any iterative traversal is managing pointers across iterations.
First, let's set up our state-of-the-world for the head (input) node.
prev
points to null
(the head has no parent), curr
points to the current node (the head), and, if there is indeed a current node, temp_next
points to the current node's child.
At each itertion, we assign the current node's next
pointer to the node at prev
(reversing the reference). We then iterate forward by pointing prev
to the current node and curr
to the original next
node.
We continue this process until a null
child is reached - at which point we can return the most recent prev
node which is our new head.
class Solution:
def reverse_list(self, node: Node) -> Node:
prev = None
curr = node
while curr:
temp_next = curr.next
curr.next = prev
prev = curr
curr = temp_next
return prev
1class Solution:
2 def reverse_list(self, node: Node) -> Node:
3 prev = None
4 curr = node
5 while curr:
6 temp_next = curr.next
7 curr.next = prev
8 prev = curr
9 curr = temp_next
10 return prev
O(n)
O(1)
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