What is Linked list and reverse the linked list iteratively.
Given a singly linked list of integers. Your task is to return the head of the reversed linked list.

For example:

The given linked list is 1 - 2 - 3 - 4- NULL. Then the reverse linked list is 4 - 3 - 2 - 1 - NULL and the head of the reversed linked list will be 4.

Follow Up :

Can you solve this problem in O(N) time and O(1) space complexity?

Input Format :

The first line of input contains an integer 'T' representing the number of test cases or queries to be processed. Then the test case follows. The only line of each test case contains the elements of the singly linked list separated by a single space and terminated by -1. Hence, -1 would never be a list element.

Output Format :

For each test case, print the given linked list in reverse order in a separate line.

Note :

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints :

1

Question

What is Linked list and reverse the linked list iteratively.

Given a singly linked list of integers. Your task is to return the head of the reversed linked list.

For example:

The given linked list is 1 -> 2 -> 3 -> 4-> NULL. Then the reverse linked list is 4 -> 3 -> 2 -> 1 -> NULL and the head of the reversed linked list will be 4.

Follow Up :

Can you solve this problem in O(N) time and O(1) space complexity?

Input Format :

The first line of input contains an integer 'T' representing the number of test cases or queries to be processed. Then the test case follows.

The only line of each test case contains the elements of the singly linked list separated by a single space and terminated by -1. Hence, -1 would never be a list element.

Output Format :

For each test case, print the given linked list in reverse order in a separate line.

Note :

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints :

1 <= T <= 5
0 <= L <= 10^5
1 <= data <= 10^9 and data != -1

Time Limit: 1 sec

Accepted Answer

Step 1: I gave all the information about linked list like how it is stored, how we iterate it, etc.
Step2 : Then i gave the iterative approach on reversing the linked list. Basically we need to take three pointers, curr, prev and next. Initialize prev as NULL, curr as head and next as NULL.Then we will iterate through the linked list. In each iteration, do following.
// Before changing next of current.store next node
next = curr->next
.
Now change next of current
// This is where actual reversing happens
curr->next = prev

// Move prev and curr one step forward
prev = curr
curr = next

I had done this question previously. Hence I was able to come up with an answer quickly. But don't worry if you don't come up with an answer in the first go. Interviewer just wanna see how you approached the problem.

Accepted Answer

Brute Force

The brute force approach is to use recursion. First, we reach the end of the Linked List recursively and at last node, we return the last node, which becomes the new head of the partially reversed Linked List. While coming back from each recursion call we add the current node in the current recursion call to the last node of the partially reversed Linked List and assign the current node to null.

Steps:

We divide the linked list of N nodes into two parts. i.e head and rest of the Linked List with (N-1) nodes.
Now recursively reverse the (N-1) nodes of Linked List and return the head of this part i.e reverseHead. After the reversal, the next node of the head will be the last node of the reversed Linked List and the head next will be pointing to this node.
But for the complete reversal of the Linked List, the head should be the last node. So, we first find the last node of this reverseHead: Linked List then we do the following:
- head.next = NULL
- temp.next = head, where the temp is the last node of the reversed linked list.
At last, return the head pointer of the reversed Linked List i.e. return reverseHead.

Space Complexity: O(n)Explanation:

O(N), where N is the total number of nodes in the Linked List.

In the worst case, extra space is required for the recursion stack.

Time Complexity: O(n^2)Explanation:

O(N ^ 2), where N is the number of nodes in the Linked List.

In the worst case, we are traversing the whole Linked List O(N) using recursion, and with each recursion call we are finding the last node of the reverse Linked List which takes O(N). Hence, the overall complexity will be O(N ^ 2).

Accepted Answer

Using Stack

Iterate through all nodes of the Linked list and push it to stack.
Then update the head pointer to the top node of the stack.
Then start popping each node from the stack and then assign the popped node next pointer to the top element of the stack.
At the last update the next pointer of the last node in the stack to null.
Now, return the head

Space Complexity: O(n)Explanation:

O(N), where N is the number of nodes in a Linked List.

In the worst case, extra space is required to store all nodes of the Linked List in the stack.

Time Complexity: O(n)Explanation:

O(N), where N is the number of nodes in the linked list.

In the worst-case, iterating the whole linked list O(N) and popping all nodes from the stack takes O(N). Hence, overall complexity will be O(N)+ O(N) i.e. O(N).

What is Linked list and reverse the linked list iteratively. Given a singly linked list of integers. Your task is to return the head of the reversed linked list. For example: The given linked list is 1 -> 2 -> 3 -> 4-> NULL. Then the reverse linked list is 4 -> 3 -> 2 -> 1 -> NULL and the head o...

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