You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

 /**

  * Definition for singly-linked list.

  * public class ListNode {

  *     int val;

  *     ListNode next;

  *     ListNode(int x) {

  *         val = x;

  *         next = null;

  *     }

  * }

  */

 public class Solution {

     public ListNode addLists(ListNode l1, ListNode l2) {

         if(l1 == null && l2 == null) {

             return null;

         }

         ListNode head = new ListNode(0);

         ListNode point = head;

         int carry = 0;

         while(l1 != null && l2!=null){

             int sum = carry + l1.val + l2.val;

             point.next = new ListNode(sum % 10);

             carry = sum / 10;

             l1 = l1.next;

             l2 = l2.next;

             point = point.next;

         }

         while(l1 != null) {

             int sum =  carry + l1.val;

             point.next = new ListNode(sum % 10);

             carry = sum /10;

             l1 = l1.next;

             point = point.next;

         }

         while(l2 != null) {

             int sum =  carry + l2.val;

             point.next = new ListNode(sum % 10);

             carry = sum /10;

             l2 = l2.next;

             point = point.next;

         }

         if (carry != 0) {

             point.next = new ListNode(carry);

         }

         return head.next;

     }

 }

 public class Solution {

     /**

      * @param l1: the first list

      * @param l2: the second list

      * @return: the sum list of l1 and l2

      */

     public ListNode addLists(ListNode l1, ListNode l2) {

         ListNode dummy = new ListNode(0);

         ListNode tail = dummy;

         int carry = 0;

         for (ListNode i = l1, j = l2; i != null || j != null; ) {

             int sum = carry;

             sum += (i != null) ? i.val : 0;

             sum += (j != null) ? j.val : 0;

             tail.next = new ListNode(sum % 10);

             tail = tail.next;

             carry = sum / 10;

             i = (i == null) ? i : i.next;

             j = (j == null) ? j : j.next;

         }

         if (carry != 0) {

             tail.next = new ListNode(carry);

         }

         return dummy.next;

     }

 }

 public class Solution {

     /*

      * @param l1: the first list

      * @param l2: the second list

      * @return: the sum list of l1 and l2

      */

     public ListNode addLists(ListNode l1, ListNode l2) {

         ListNode root = new ListNode(-1);

         ListNode result = root;

         int carry = 0;        

         while( l1 != null || l2 != null || carry == 1){

             int value = 0;

             if(l1 != null){

                 value += l1.val;

                 l1 = l1.next;

             }

             if( l2 != null){

                 value += l2.val;

                 l2 = l2.next;

             }

             value += carry;

             root.next = new ListNode(value % 10);

             carry = value / 10;

             root = root.next;

         }

         return result.next;

     }

 }

 class Solution {

 public:

     ListNode* addLists(ListNode* l1, ListNode* l2) {

         ListNode *head = new ListNode();

         ListNode *ptr = head;

         int carry = ;

         while (true) {

             if (l1 != NULL) {

                 carry += l1->val;

                 l1 = l1->next;

             }

             if (l2 != NULL) {

                 carry += l2->val;

                 l2 = l2->next;

             }

             ptr->val = carry % ;

             carry /= ;

             // 当两个表非空或者仍有进位时需要继续运算，否则退出循环

             if (l1 != NULL || l2 != NULL || carry != ) {

                 ptr = (ptr->next = new ListNode());

             } else {

                 break;

             }

         }

         return head;

     }

 };