Reversing a singly linked list is a fundamental problem in computer science and forms a part of many coding interviews. In this blog post, we'll delve into the steps involved in reversing a singly linked list using Python. We'll also discuss the provided code snippet, explain each step, and analyze its time and space complexity.

**Understanding the Problem:**

Given the head of a singly linked list, our task is to reverse the list and return the reversed version. For example, if the input linked list is `[1, 2, 3, 4, 5]`

, the reversed list should be `[5, 4, 3, 2, 1]`

.

**Approach:**

We'll utilize a simple iterative approach to reverse the linked list. The idea is to traverse the list while maintaining three pointers: `prev`

, `curr`

, and `nextNode`

.

**Explanation of the Code:**

Let's break down the provided Python code snippet:

```
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
prev, curr = None, head
while curr:
nextNode = curr.next # Store the next node
curr.next = prev # Reverse the pointer
prev = curr # Move prev to curr
curr = nextNode # Move curr to the next node
return prev
```

We define a class `Solution`

with a method `reverseList`

, which takes the head of the linked list as input and returns the reversed list.

- Initially, we set
`prev`

to`None`

and`curr`

to the head of the linked list. - We enter a
`while`

loop that iterates until`curr`

becomes`None`

, indicating the end of the list. - Inside the loop, we store the next node of
`curr`

in`nextNode`

. - We then reverse the link of
`curr`

by pointing its`next`

to`prev`

. - After reversing the link, we move
`prev`

to`curr`

and`curr`

to`nextNode`

, progressing to the next nodes. - Once the loop ends,
`prev`

points to the new head of the reversed list, so we return`prev`

.

**Time Complexity:**

The time complexity of this solution is O(n), where n is the number of nodes in the linked list. This is because we traverse the entire list once.

**Space Complexity:**

The space complexity is O(1) since we use a constant amount of extra space regardless of the input size. We only use a few extra pointers to keep track of the nodes being processed.

**Conclusion:**

Reversing a singly linked list is a common problem in coding interviews and real-world applications. By understanding the iterative approach and the provided code snippet, you can effectively tackle this problem. The time and space complexities of the solution demonstrate its efficiency, making it a practical solution for reversing linked lists in Python.

Happy coding!