Reverse a singly linked list.

public class Solution {
    /**
     * @param head: The head of linked list.
     * @return: The new head of reversed linked list.
     */
    public ListNode reverse(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        ListNode rev = null;
        while (head != null) {
            ListNode next = head.next;
            head.next = rev;
            rev = head;
            head = next;
        }
        return rev;
    }
    //recursiton
    public ListNode reverseList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        ListNode rev = reverseList(head.next);
        //head.next is the last element in rev, we connect it to the first element, and clear the first element.
        head.next.next = head;
        head.next = null;
        return rev;
    }
}

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """

        if not head:
            return

        rev = None
        while head:
            next_node = head.next
            head.next = rev
            rev = head
            head = next_node
        return rev

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* dummy = NULL;
        while (head != NULL) {
            ListNode* next1 = head->next;
            head->next = dummy;
            dummy = head;
            head = next1;
        }
        return dummy;
    }
    };