简单明了的解题思路

首先题目保存的是正序的数字，而计算两数相加一定要从个位也就是链表的尾部进行计算，并逐步进位。而单链表如果从后向前遍历明显不是很方便，自然想到后进先出的栈，然后就是简单的实现就可。

• 初始化栈
• 出栈并计算
• 处理剩余值
• 处理最后进位

为了方便对每位的计算结果进行头插，引入了一个辅助头，下面是自己的垃圾代码，希望对你有所帮助（小白第一次写，有问题请在评论区指出）

import java.util.*;

/*
 * 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
        Deque<Integer> keep1 = new LinkedList<Integer>();
        Deque<Integer> keep2 = new LinkedList<Integer>();
        ListNode head = new ListNode(-1);
        //初始化栈
        while(head1!=null){
            keep1.offerLast(head1.val);
            head1 = head1.next;
        }
        while(head2!=null){
            keep2.offerLast(head2.val);
            head2 = head2.next;
        }
        //开始计算，count保存进位
        int count = 0;
        while(!keep1.isEmpty() && !keep2.isEmpty()){
            int a = keep1.pollLast();
            int b = keep2.pollLast();
            int sum = a+b+count;
            count = sum / 10;
            ListNode tmp = new ListNode(sum%10);
            tmp.next = head.next;
            head.next = tmp;
        }
        //余值计算
        Deque<Integer> keep = keep1.isEmpty()?keep2:keep1;
        while(!keep.isEmpty()){
            int t = keep.pollLast();
            int sum = t+count;
            count = sum / 10;
            ListNode tmp = new ListNode(sum%10);
            tmp.next = head.next;
            head.next = tmp;
        }
        //最后的进位判断
        if(count!=0){
            ListNode tmp = new ListNode(1);
            tmp.next = head.next;
            head.next = tmp;
        }
        
        return head.next;
    }
}