Description:

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

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

思路：

因为两个数字都是倒序储存的，所以直接送链表的开头逐个相加即可，但是需要注意的是，当整个循环结束后，可能进位不是0，所以要把进位加到结果中去，这是一个比较重要的临界条件。

Solution:

public static ListNode add_tow_numbers(ListNode l1, ListNode l2){
  ListNode head = new ListNode(0);
  ListNode cur = head;
  int carry = 0;
  while(l1 != null || l2 != null){
    int x = (l1 != null) ? l1.val : 0;  //缺少的位用零补上
    int y = (l2 != null) ? l2.val : 0;
    int sum = x + y + carry;
    carry = sum / 10;
    cur.next = new ListNode(sum %= 10);
    cur = cur.next;
    if(l1 != null) l1 = l1.next;  //先判断是不是null，否则可能出现NullPointerException
    if(l2 != null) l2 = l2.next;
  }
  if(carry != 0){   //如果还有进位就加到结果的最后
    cur.next = new ListNode(carry);
  }
  return head.next;
}