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;

    }
    
}