Lintcode-easy-add numbers

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

Example

Given 7->1->6 + 5->9->2 . That's, 617 + 295 .

Return 2->1->9 . That's 912 .

Given 3->1->5 5->9->2 and, return 8->0->8 .

There's not much to say about this question, and it's about the same as putting two binary numbers together.

/*** Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * ListNode (int       X) {* val = x; * next = NULL; *     } * } */ Public classSolution {/**     * @paraml1:the First list *@paraml2:the Second list *@return: The sum list of L1 and L2*/     PublicListNode addlists (listnode L1, ListNode L2) {//Write your code here                if(L1 = =NULL)            returnL2; if(L2 = =NULL)            returnL1; ListNode Fakehead=NewListNode (0); ListNode P=Fakehead; ListNode P1=L1; ListNode P2=L2; intCarry = 0;  while(P1! =NULL|| P2! =NULL){            intNUM1 = 0; intnum2 = 0; if(P1! =NULL) {NUM1=P1.val; P1=P1.next; }            if(P2! =NULL) {num2=P2.val; P2=P2.next; } P.next=NewListNode ((NUM1 + num2 + carry)% 10); P=P.next; Carry= (NUM1 + num2 + carry)/10; }                if(Carry = = 1) {P.next=NewListNode (1); P=P.next; }                returnFakehead.next; }}

Lintcode-easy-add numbers