Reverse Linked List

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

Constraints:

    The number of nodes in the list is the range [0, 5000].
    -5000 <= Node.val <= 5000

Idea

Using a dummy head could make things easier.

For each node in the list:

Set node as dummy's successer

Set dummy's successor to this node

Code

1. Iterative solution:

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode dummy = new ListNode(0);
        ListNode cur = head;
        while (cur != null) {
            ListNode next = cur.next;
            cur.next = dummy.next;
            dummy.next = cur;
            cur = next;
        }
        
        return dummy.next;
    }
}

• Time: O(n) as we have to traverse all the nodes in the linked list.
• Space: O(1) as we use a dummy node as the fake head of the reversed list.

2. Recursive solution

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode dummy = new ListNode(0);
        reverse(head, dummy);
        return dummy.next;
    }
    
    private void reverse(ListNode head, ListNode dummy) {
        if (head == null) {
            return;
        }
        
        ListNode next = head.next;
        head.next = dummy.next;
        dummy.next = head;
        reverse(next, dummy);
    }
}

• Time: O(n) as we have to traverse all the nodes in the linked list.
• Space: O(n) as the recursive method will invoked against each node in the list.

Similar question:

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