Merge Two Sorted Lists — Dummy Head Merge

// C
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode dummy = {0, NULL}, *cur = &dummy;
    while (l1 && l2) {
        if (l1->val <= l2->val) { cur->next = l1; l1 = l1->next; }
        else { cur->next = l2; l2 = l2->next; }
        cur = cur->next;
    }
    cur->next = l1 ? l1 : l2;
    return dummy.next;
}

// C++
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
    ListNode dummy(0); ListNode* cur = &dummy;
    while (l1 && l2) {
        if (l1->val <= l2->val) { cur->next = l1; l1 = l1->next; }
        else { cur->next = l2; l2 = l2->next; }
        cur = cur->next;
    }
    cur->next = l1 ? l1 : l2;
    return dummy.next;
}

// Java
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0), 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;
}

# Python
def mergeTwoLists(l1, l2):
    dummy = cur = ListNode(0)
    while l1 and l2:
        if l1.val <= l2.val: cur.next = l1; l1 = l1.next
        else: cur.next = l2; l2 = l2.next
        cur = cur.next
    cur.next = l1 or l2
    return dummy.next