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Reverse Linked List

Certification: Basic Level Accuracy: 0% Submissions: 0 Points: 5

Write a C program to reverse a singly linked list. Given a pointer to the head of a linked list, return a pointer to the head of the reversed list. The function should take the head of the original list as input and return the head of the reversed list.

Example 1
    
  
  • Input: 1->2->3->4->5->NULL
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  • Output: 5->4->3->2->1->NULL
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  • Explanation: 
Step 1: We start with a linked list 1->2->3->4->5->NULL.
Step 2: We need to reverse the direction of each pointer.
Step 3: First, 1->NULL (previous is NULL, current is 1, next is 2).
Step 4: Then, 2->1->NULL (previous is 1, current is 2, next is 3).
Step 5: Then, 3->2->1->NULL (previous is 2, current is 3, next is 4).
Step 6: Then, 4->3->2->1->NULL (previous is 3, current is 4, next is 5).
Step 7: Finally, 5->4->3->2->1->NULL (previous is 4, current is 5, next is NULL).
Step 8: The new head of the reversed list is 5.
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Example 2
    
  
  • Input: 1->NULL
    • 
  
  • Output: 1->NULL
    • 
  
  • Explanation:
Step 1: We start with a linked list containing only one node: 1->NULL.
Step 2: The reverse of a single-node list is the list itself.
Step 3: So the output is also 1->NULL.
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Constraints
    
  
  • The number of nodes in the list is in the range [0, 5000]
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  • The values of the nodes are in the range [-5000, 5000]
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  • You must reverse the list in-place with O(1) extra memory
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  • Time Complexity: O(n) where n is the number of nodes in the linked list
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  • Space Complexity: O(1)
    •
Linked ListGoldman SachsZomato
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Hint
Solution

Solution Hints

    
  
  • Use three pointers to track the previous, current, and next nodes
    • 
  
  • Initialize previous as NULL, current as head
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  • Iterate through the list, at each step:
   - Save the next node before modifying the current node
   - Reverse the current node's pointer to point to the previous node
   - Move previous and current pointers one step forward
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  • After the iteration, previous will be pointing to the new head of the reversed list
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  • Handle edge cases like an empty list or a list with only one node
    •

Steps to solve by this approach:

 Step 1: Initialize three pointers: prev = NULL, current = head, next = NULL

 Step 2: Iterate through the linked list using a while loop until current becomes NULL

 Step 3: Inside the loop, first store the next node (next = current->next)

 Step 4: Reverse the current node's pointer to point to the previous node (current->next = prev)

 Step 5: Move prev and current pointers one step forward (prev = current, current = next)

 Step 6: After the loop ends, prev will be pointing to the new head of the reversed list

 Step 7: Return prev as the new head of the reversed list

Submitted Code :