思路:
1.两数相加肯定是先个位相加,然后进位
2.反转链表,链表头就是个位数字了,然后相加,相加后,商的值就是当前节点值,余数参数下一个节点计算进位。(因为可能最后一个节点是0,所以反转后的到的节点,如果头节点是0的话,就要去掉。看你怎么写了,因为在所有节点计算完后,余数可能为0,也可能非0,需要新建个节点来存储,如果是0的话,那计算完,反转后,最高位不能是0)
3.当然,没有链表是以0开头的节点,0开头的正数怎么可能,如果你担心的话,那可以遍历头节点去除值为0的,知道下一个值不为0的节点。


/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    public ListNode addInList (ListNode head1, ListNode head2) {
        // write code here
        if(head1 == null  && head2 == null){
            return null;
        }
        if(head1 == null){
            return head2;
        }
        if(head2 == null){
            return head1;
        }
        head1 = reverse(head1);
        head2 = reverse(head2);
        //开始从最右侧 个位相加了
        int addition = 0;
        ListNode head = new ListNode(-1);
        ListNode current = head;
        while(head1 != null || head2 != null){
            int a = 0, b = 0;
            if(head1 != null){
                a = head1.val;
                head1 = head1.next;
            }
            if(head2 != null){
                b = head2.val;
                head2 = head2.next;
            }
            int sum = addition + a + b;
            current.val = sum % 10;
            addition = sum / 10;
            ListNode next = new ListNode(-1);
            current.next = next;
            current = next;
            
        }
        current.val = addition;
        head = reverse(head);
        if(head.val == 0){
            head = head.next;
        }
        return head;
    }
     private ListNode reverse(ListNode head){
        ListNode cur = head;
        ListNode pre = null;
        while(cur != null){
            ListNode temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }
}