Leetcode ADD Numbers

In vs2010, it is no problem to have the pointer declared without assigning a value and pail null. However, if you do not assign a value on the Leetcode, it must be initialized to null, or re.

/** * Definition for singly-linked list.
 * struct ListNode {* int val;
 * ListNode *next;
 * ListNode (int x): Val (x), Next (NULL) {}};
    	*/class Solution {Public:listnode *addtwonumbers (ListNode *l1, ListNode *l2) {ListNode *res = NULL;
    	if (!L1) {res = L2;return res;
    	} if (!L2) {res = L1;return res;
    	} ListNode *p = NULL;
    	int carry = 0, tmp;
    		while (L1 && L2) {tmp = L1->val + l2->val + carry;
    		carry = TMP/10;
    		TMP%= 10;
    			if (!p) {p = new ListNode (TMP);
    		res = p;
    			}else{ListNode *ntmp = new ListNode (TMP);
    			P->next = ntmp;
    		p = ntmp;
    		} L1 = l1->next;
    	L2 = l2->next;
    		} while (L1) {tmp = l1->val + carry;
    		carry = TMP/10;
    		TMP%= 10;
    			if (!p) {p = new ListNode (TMP);
    		res = p;
    			}else{ListNode *ntmp = new ListNode (TMP);
    			P->next = ntmp; p = Ntmp
    	} L1 = l1->next;
    		} while (L2) {tmp = l2->val + carry;
    		carry = TMP/10;
    		TMP%= 10;
    			if (!p) {p = new ListNode (TMP);
    		res = p;
    			}else{ListNode *ntmp = new ListNode (TMP);
    			P->next = ntmp;
    		p = ntmp;
    	} L2 = l2->next;
    		} if (carry>0) {ListNode *ntmp = new ListNode (carry);
    	P->next = ntmp;
    } return res; }
};

However, there are some more concise wording:

Public ListNode addtwonumbers (listnode L1, ListNode L2) {
        ListNode current = new ListNode (0);
        ListNode head = current;

        int shift = 0;
        Do
        {
            current.val = ((l1!=null) l1.val:0) + ((l2!=null)? l2.val:0) + Shift;
            shift = CURRENT.VAL/10;
            current.val%=10;
            L1 = (l1!=null)? L1.next:null;
            L2 = (l2!=null)? L2.next:null;
            if ((l1==null) && (l2==null)) {break;}

            Current.next = new ListNode (0);
            current = Current.next;
        } while (L1!=null | | l2!=null);
        if (shift>0) {current.next = new ListNode (1);}
        return head;
    }