Problem Statement:
Explanation:
- You are given the head of a singly linked list.
- You need to reverse the linked list and return its head.
- A linked list is a sequence of elements, where each element points to the next one.
Sample Data:
- Input: head = 1 -> 2 -> 3 -> None
- Output: 3 -> 2 -> 1 -> None
Questions to be Asked to the Interviewer:
- What should be returned if the linked list is empty?
- Are there any constraints on the length of the linked list?
Edge Cases for this Problem:
- If the linked list is empty, the result should also be an empty linked list.
Available Approaches to Solve this Problem:
- Iterative Approach: Use three pointers to reverse the direction of the links between the nodes.
- Recursive Approach: Recursively reverse the rest of the list and then change the next pointer of the current node.
My Approach to Solve this Problem:
- I will use the iterative approach, reversing the links between the nodes as I traverse the list.
Classification:
- Algorithm: Iteration
- Data Structure: Linked List
class ListNode:
? ? def __init__(self, val=0, next=None):
? ? ? ? self.val = val
? ? ? ? self.next = next
def reverseList(head):
? ? prev = None
? ? current = head
? ??
? ? while current:
? ? ? ? next_node = current.next
? ? ? ? current.next = prev
? ? ? ? prev = current
? ? ? ? current = next_node
? ??
? ? return prev
Explain the Solution to a Non-Programmer:
- Think of a linked list as a chain of paper clips, each attached to the next.
- We want to reverse the direction of the chain, so the last paper clip becomes the first, and so on.
- We do this by detaching and reattaching the paper clips in reverse order, starting from the first paper clip.
Code in Detail:
- Line 1-3: Define a class ListNode to represent each node (paper clip) in the linked list.
- Line 5: Define the reverseList function that takes the head of the linked list.
- Line 6: Initialize a prev pointer as None to keep track of the previous node.
- Line 7: Initialize a current pointer starting at the head to keep track of the current node.
- Lines 9-14: Loop through the linked list:Store the next node temporarily (next_node).
- Reverse the direction of the current link by making the current node's next pointer point to the previous node.
- Move the prev pointer to the current node and current pointer to the next node (next_node).
- Line 16: Return the new head of the reversed linked list, which is where the prev pointer ends up.
Pattern of this Problem:
The pattern for reversing a linked list involves the use of pointers. Understanding how pointers are manipulated is key to solving the problem.
Big O Notation:
- Time Complexity: O(n)-O(n), where n is the length of the linked list. We traverse the list once.
- Space Complexity: O(1)-->O(1), as we are only using a constant amount of extra space.
Points to Remember to Solve this Problem in the Future:
- Reversing a linked list requires careful manipulation of pointers.
- Utilize temporary variables to avoid losing references.
- Be mindful of edge cases, such as an empty linked list.
Code Line by Line with Sample Data:
- Input: head = 1 -> 2 -> 3 -> None
- Line 6: prev is set to None.
- Line 7: current is set to head (1).
- Line 9: Loop begins.Line 10: next_node is set to 2.
- Line 11: 1's next is set to prev (None), reversing the link.
- Line 12: prev is set to 1.
- Line 13: current is set to 2.
- Next iterations: Continues until current is None.
- Line 16: Returns 3 -> 2 -> 1 -> None.
Key Syntax Used in this Solution:
- Linked List Manipulation: Working with pointers, such as next_node = current.next.
- While Loop: while current:.
- Class Definition for a Node: class ListNode:.
