In this series of posts, I will discuss coding questions on the `LinkedList`

Data structure.

The posts in this series will be organized in the following way,

- Question Link โ
- Possible Explanation ๐
- Documented C++ Code ๐งน
- Time and Space Complexity Analysis โ๐

## The Question

Given theย `head`

ย of a singly linked list, reverse the list, and return the reversed list.

https://leetcode.com/problems/reverse-linked-list/

๐ก Give yourself at least 15-20 mins to figure out the solution :)

## Explanation

Its bit **tricky** so pay your atmost attention,

The idea is to make *current* node's successor( `curโnext`

) point to *current* and there by reversing the list. (Here, we are under assumption that the list after current node is already reversed ).

Lastly, we will return the last node's pointer as the new `head`

of our reversed linkedlist.

See the recursion animation to make things more clear,

## C++ Code

### Definition of Linked List

```
//Definition for singly-linked list.
struct ListNode
{
int val;
ListNode *next;
ListNode() : val(0), next(nullptr) {}
ListNode(int x) : val(x), next(nullptr) {}
ListNode(int x, ListNode *next) : val(x), next(next) {}
};
```

### Solution

```
ListNode *reverseList(ListNode *head)
{
/*
*Approach
-Let's say list is n0 -> n1 ->... "nk" -> nk+1 -> ...n->null
-when we are on nk, we assume nk+1 <-...n i.e. the list ahead is reversed
*/
//* When 1. list contains 0 or 1 node,
//* 2. head is pointing last node
if (head == nullptr || head->next == nullptr)
return head;
ListNode *last = reverseList(head->next);
//note: To make sure this works, we need to make sure:
//> head->next and head exits (base condition)
head->next->next = head; //! crux
head->next = nullptr;
return last;
}
```

## Complexity Analysis

### Time Complexity: O(n)

We will visit every node once.

### Space Complexity: O(n)

The size of recursion stack as we will go **n** levels deep.

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