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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */

Why dummy and cur ?

cur is set to create the whole res list; dummy is still at the beginning of the res list; returning the head of the list.
所以很多时候是创建了两个节点，一个dummy一直都在最前面最后返回的也会是这个node，另一个cur作为处理过程中用的node, 并不影响dummy

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class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null && l2 == null) return null;
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                cur.next = l1;
                l1 = l1.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        if (l1 != null) cur.next = l1;
        if (l2 != null) cur.next = l2;
        return dummy.next;
    }
}

or shorter:

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public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1), cur = dummy;
        while (l1!=null && l2!=null) {
            if (l1.val < l2.val) {
                cur.next = l1;
                l1 = l1.next;
            } else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        cur.next = l1!=null ? l1 : l2;
        return dummy.next;
    }

essentially they are the same