Table Of Contents
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- Problem Statement
- Recursive Approach
- Implementation Of Recursive Approach
- Iterative Approach
- Dry Run of the Iterative Approach
- Implementation of Iterative Approach
- Practice Questions
- Frequently Asked Questions
- Q.1: What is the time complexity of reversing a linked list?
- Q.2: Can we use Stack to reverse the Linked List?
- Q.3: Will the above approaches change the address of the node?
- Additional Resources
Problem Statement
Given a pointer to the head node of a linked list, the task is to reverse the linked list. We need to switch the list by changing the links between nodes.
Example:
- Input: [1,2,3,4,5,NULL]
- Output: [5,4,3,2,1,NULL]
- Explanation:
- Input: [3,4,5]
- Output: [5,4,3]
- Explanation:
Recursive Approach
The recursive approach to reverse a linked list is simple, we have to divide the linked lists into two parts and i.e. first node and the rest of the linked list, and then call the recursion for the other part by maintaining the connection.
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Implementation Of Recursive Approach
C++ Implementation
ListNode* reverseList(ListNode* head) {
if(!head || !(head->next)) return head;
auto res = reverseList(head->next);
head->next->next = head;
head->next = NULL;
return res;
}Java Implementation
static class Node {
int data;
Node next;
Node(int d) {
data = d;
next = null;
}
}
static Node reverse(Node head) {
if (head == null || head.next == null)
return head;
Node rest = reverse(head.next);
head.next.next = head;
head.next = null;
return rest;
}Python Implementation
def reverse(self, head):
# If head is empty or has reached the list end
if head is None or head.next is None:
return head
# Reverse the rest list
rest = self.reverse(head.next)
# Put first element at the end
head.next.next = head
head.next = None
# Fix the header pointer
return rest- Time complexity: O(N), Where N is the size of the linked list.
- Space complexity: O(1)
Iterative Approach
- We will use 3 variables: prevNode, head, and nextNode.
- prevNode to NULL
- nextNode can stay empty;
- Then we will continue our loop until our current maidenhead pointer is truthy (ie: not NULL), which implies that we reached the end of the linked list
- During our loop, we first of all update nextNode so that it acquires its namesake value, as head->next.
- Finally, we update the head with the value we stored in nextNode.
- And finally, we go on with the loop until we can. After the loop, we return prevNode.
Dry Run of the Iterative Approach
Implementation of Iterative Approach
Reverse A Linked List In C++
ListNode* reverseList(ListNode* head) {
ListNode *prev = NULL, *cur=head, *tmp;
while(cur){
tmp = cur->next;
cur->next = prev;
prev = cur;
cur = tmp;
}
return prev;
}Reverse A Linked List In Java
static class Node {
int data;
Node next;
Node(int d) {
data = d;
next = null;
}
}
/* Function to reverse the linked list */
Node reverse(Node node) {
Node prev = null;
Node current = node;
Node next = null;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
node = prev;
return node;
}Reverse A Linked List In Python
def reverse(self): prev = None current = self.head while (current is not None): next = current.next current.next = prev prev = current current = next self.head = prev
- Time complexity: O(N), Where N is the size of the array.
- Space complexity: O(1)
Practice Questions
Frequently Asked Questions
Q.1: What is the time complexity of reversing a linked list?
Ans: It is linear in time i.e. O(n) where n is the size of the linked list.
Q.2: Can we use Stack to reverse the Linked List?
Ans: Yes, but the space complexity will increase to O(N).
Q.3: Will the above approaches change the address of the node?
Ans: No, we should try not to change the address of the node.
