![]() ![]() This approach has a time complexity of O(n) but uses O(n) space due to the stack. # Creating a new linked list from the stack We then pop elements from the stack, which will be in reversed order, and create a new linked list. We can traverse the original linked list and push each node onto a stack. We can also reverse a linked list by using a stack. The recursive approach also has a time complexity of O(n), but the space complexity is O(n) due to the recursive call stack. In this method, we recursively reverse the sub-list leaving the first element and then reverse the pointer of the first element. Approach 2: Recursive MethodĪnother approach to reverse a linked list is to use recursion. This approach has a time complexity of O(n) and a space complexity of O(1). We traverse the list and reverse the pointers iteratively. The most common method to reverse a linked list involves using an iterative approach where we use three pointers: prev, curr, and next. Self.next = next Approach 1: Iterative Method The first node is called the head of the list, and the last node points to NULL, indicating the end of the list. Understanding Linked ListsĪ singly linked list consists of nodes, where each node contains a data element and a pointer to the next node in the list. The task is to reverse the pointers so that the last element points to the second last, and so on. The Reverse Linked List problem (LeetCode #206) is defined as follows:Įxample: Input: 1 -> 2 -> 3 -> 4 -> 5 -> NULL Output: 5 -> 4 -> 3 -> 2 -> 1 -> NULLĪ linked list is a data structure where each element points to the next one. This article aims to provide an in-depth look at the Reverse Linked List problem on LeetCode, various approaches to solve it, and Python code implementations. Being proficient in manipulating linked lists is essential for anyone delving into computer science and programming. ![]() ![]() Reversing a linked list is a classic algorithmic problem often asked in coding interviews and present on competitive programming platforms like LeetCode. ![]()
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