Reverse a Linked List — Iterative & Recursive (Python, C, JS)
Problem Statement
Given the head of a singly linked list, reverse the list and return the reversed list's head.
Example
Input: head = [1 -> 2 -> 3 -> 4 -> 5]
Output: [5 -> 4 -> 3 -> 2 -> 1]
Time Complexity
O(n)
Space Complexity
O(1)
The Intuition
Three-pointer iterative — maintain prev (None), curr, and temp_next. Flip curr.next = prev, then advance both pointers forward.
Solution Code
def reverseList(head):
prev = None
curr = head
while curr:
nxt = curr.next
curr.next = prev
prev = curr
curr = nxt
return prevstruct ListNode* reverseList(struct ListNode* head) {
struct ListNode *prev=NULL, *curr=head, *nxt=NULL;
while (curr) {
nxt = curr->next;
curr->next = prev;
prev = curr;
curr = nxt;
}
return prev;
}function reverseList(head) {
let prev = null, curr = head;
while (curr) {
const nxt = curr.next;
curr.next = prev;
prev = curr;
curr = nxt;
}
return prev;
}Frequently Asked Questions
Q: Why store curr.next in a temp variable?
Because you overwrite curr.next when flipping the pointer, losing access to the rest of the list without that temporary save.
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