Reverse a Singly Linked List

easy linked list pointers

Problem

Given the head of a singly linked list, reverse the direction of every next pointer in place and return the new head. Use only constant extra space.

Input1 → 2 → 3 → null

Output3 → 2 → 1 → null

node values (comma-separated)

def reverse_list(head):
    prev, curr = None, head
    while curr is not None:
        nxt = curr.next
        curr.next = prev
        prev = curr
        curr = nxt
    return prev

function reverseList(head) {
  let prev = null;
  let curr = head;
  while (curr !== null) {
    const next = curr.next;
    curr.next = prev;
    prev = curr;
    curr = next;
  }
  return prev;
}

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
}

ListNode* reverseList(ListNode* head) {
    ListNode* prev = nullptr;
    ListNode* curr = head;
    while (curr) {
        ListNode* next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}

Time: O(n) Space: O(1)