2 Add Two Numbers

• 题目描述：
  给两个数字组成的链表，计算它们从低位到高位的和，最后输出一个和的链表头结点
  Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
  Output: 7 -> 0 -> 8

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int carry =0;
        ListNode newHead = new ListNode(0);
        ListNode p1 = l1, p2 = l2, p3=newHead;
 
        while(p1 != null || p2 != null){
            if(p1 != null){
                carry += p1.val;
                p1 = p1.next;
            }
 
            if(p2 != null){
                carry += p2.val;
                p2 = p2.next;
            }
 
            p3.next = new ListNode(carry%10);
            p3 = p3.next;
            carry /= 10;
        }
 // 最后高位判定加一
        if(carry==1) 
            p3.next=new ListNode(1);
 
        return newHead.next;
    }
}