# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    '''
    Time Complexity: O(n + m)
    Space Complexity: O(1)
    where, n and m are the number of nodes in list1 and list2 respectively.
    '''
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        dummy = ListNode()
        node = dummy
        while list1 and list2:
            if list1.val < list2.val:
                temp = list1.next
                list1.next = None
                node.next = list1
                node = list1
                list1 = temp
            else:
                temp = list2.next
                list2.next = None
                node.next = list2
                node = list2
                list2 = temp
        node.next = list1 if list1 else list2
        return dummy.next

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
    ListNode dummy = new ListNode();
    ListNode cur = dummy;
    while (list1 != null && list2 != null) {
        if (list1.val < list2.val) {
            cur.next = list1;
            list1 = list1.next;
        } else {
            cur.next = list2;
            list2 = list2.next;
        }
        cur = cur.next;
    }
    if (list1 != null) {
        cur.next = list1;
    } else {
        cur.next = list2;
    }
    return dummy.next;
}