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Reverse a Linked List

Certification: Basic Level Accuracy: 100% Submissions: 2 Points: 10

Write a C# program to reverse a singly linked list. Given a pointer to the head node of a linked list, the task is to reverse the linked list.

Example 1
    
  
  • Input: head = [1,2,3,4,5]
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  • Output: [5,4,3,2,1]
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  • Explanation: 
      
      
    • Step 1: Initialize three pointers: prev as null, current as head, and next as null.
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    • Step 2: Iterate through the list 1->2->3->4->5
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    • Step 3: For each node, save the next node, reverse the current node's pointer, move pointers one position ahead.
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    • Step 4: Return the new head of the reversed list, which is 5.
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Example 2
    
  
  • Input: head = [1,2]
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  • Output: [2,1]
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  • Explanation: 
      
      
    • Step 1: Initialize three pointers: prev as null, current as head, and next as null.
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    • Step 2: Iterate through the list 1->2
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    • Step 3: Save node 2 as next, point node 1 to null, move prev to node 1, move current to node 2.
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    • Step 4: Save null as next, point node 2 to node 1, move prev to node 2, current becomes null.
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    • Step 5: Return the new head of the reversed list, which is 2.
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Constraints
    
  
  • 0 ≤ Number of nodes ≤ 5000
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  • -5000 ≤ Node.val ≤ 5000
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  • Time Complexity: O(n) where n is the number of nodes in the list
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  • Space Complexity: O(1)
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Linked ListPointers and ReferencesWiproDropbox
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Hint
Solution

Solution Hints

    
  
  • Use three pointers to keep track of the current node and its previous and next nodes.
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  • Iterate through the list, changing each node's next pointer to point to the previous node.
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  • Return the new head of the reversed list.
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  • Be careful when handling edge cases like an empty list or a list with only one node.
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  • Consider both iterative and recursive approaches.
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Steps to solve by this approach:

 Step 1: Initialize three pointers: prev as null, current as head, and next as null.
 Step 2: Iterate through the list until current becomes null.
 Step 3: Inside the loop, save the next node before modifying the current node.
 Step 4: Reverse the current node's pointer to point to the previous node.
 Step 5: Move the prev and current pointers one step forward for the next iteration.

Submitted Code :