2. Add Two Numbers(2个链表相加)

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You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

20180223

 class Solution {
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
         ListNode fakehead = new ListNode(0);
         ListNode prev = fakehead;
         int carry = 0;
         for(ListNode p1=l1, p2 = l2;
            p1!=null || p2!=null;
            p1=(p1==null?null:p1.next),p2=(p2==null?null:p2.next)
            ){
                int p1val = p1==null?0:p1.val;
                int p2val = p2==null?0:p2.val;
                int val = p1val+p2val+carry;
                carry = val/10;
                val = val%10;
                ListNode temp = new ListNode(val);
                prev.next = temp;
                prev = prev.next;
            }
             if(carry>0)
                 prev.next = new ListNode(carry);
             return fakehead.next;
     }
 }

 class Solution {
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        // ListNode cur =new ListNode(0);
         ListNode prev= new ListNode(0);
         ListNode head = prev;
         int pval;
         int jinwei=0;
         while(l1!=null ||l2!=null || jinwei!= 0 ){
             pval  = ((l2 == null) ? 0 : l2.val) + ((l1 == null) ? 0 : l1.val) + jinwei;
             if(pval>9) {   jinwei =1;pval=pval-10;    }
             else jinwei=0;
             ListNode cur = new ListNode(pval);
             prev.next =cur;
             prev = cur;
             l1 = (l1 == null) ? l1 : l1.next;
             l2 = (l2 == null) ? l2 : l2.next;

         }
         return head.next;

     }

 }