LeetCode Add Two Numbers 解决代码

暴露出了自己做链表题的很多缺点，希望自己以后能多引以为戒

Problem:

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.
Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.class Solution {

   /*传入的参数为两个链表 l1 l2，最终结果保存在l1中

*用 l3指向l1，最后返回l3

*/ 

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

    int flag = 0;
int num = 1;
ListNode l3 = new ListNode(0);//这个l3必须初始化，否则多次调用程序可能结果不对
ListNode watch;
l3 = l1;  
watch = l1;                                                    //看门狗，指向存储结果链表的最后一个节点   
while (l1 != null && l2 != null) {                   //l1.next!=null的话会出现什么结果要如何处理
if (l1.val + l2.val + flag >= 10) {
l1.val = (l1.val + l2.val + flag) % 10;
flag = 1;
} else {
l1.val = l1.val + l2.val + flag;
flag = 0;
}
l1 = l1.next;
l2 = l2.next;
if (num == 1)
num--;
else {
watch = watch.next;
}
}
if (l1 == null && l2 == null) {                                   //至少一个为空退出循环有三种可能，二者都结束或其中只有

if (flag != 0) {
// 一个为空
ListNode tListNode = new ListNode(flag);
watch.next = tListNode;
}
}
if (l1 != null) {                                                           
//l1不为空，l1直接指向自己的后继
while (l1.next != null) {                                           // 到达最后节点，需要特别考虑是否需要进位
if (l1.val + flag >= 10) {
l1.val = (l1.val + flag) % 10;
flag = 1;
} else {
l1.val = l1.val + flag;
flag = 0;
}
l1 = l1.next;
}
if (l1.val + flag >= 10) {        //最后一个节点发生了进位
l1.val = (l1.val + flag) % 10;
flag = 1;
}else    {                      

            l1.val = l1.val + flag;
flag = 0;

            }

            if(flag != 0){                                   //进位则创建新节点

                ListNode temp = new ListNode(flag);
    l1.next = temp;

            }
}

if (l2 != null) {
watch.next = l2;
watch = watch.next;
while (watch.next != null) {
if (watch.val + flag >= 10) {
watch.val = (watch.val + flag) % 10;
flag = 1;
} else {
watch.val = watch.val + flag;
flag = 0;
}
watch = watch.next;
}
if (watch.val + flag >= 10) {
watch.val = (watch.val + flag) % 10;
flag = 1;
}else{

                watch.val = watch.val+flag;flag=0;

            }

            if(flag!=0){

            ListNode temp = new ListNode(flag);
    watch.next = temp;   

            }

}
return l3;

    

    }

}