445. Add Two Numbers II【Medium】【两个链表求和】

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        Stack<Integer> stack1 = new Stack<>();
        Stack<Integer> stack2 = new Stack<>();

        while (l1 != null) {
            stack1.push(l1.val);
            l1 = l1.next;
        }
        while (l2 != null) {
            stack2.push(l2.val);
            l2 = l2.next;
        }
        //node的下一个结点为head
        ListNode head = new ListNode(0); //head记录结果
        while (!stack1.isEmpty() || !stack2.isEmpty()) {
            if (!stack1.isEmpty()) head.val += stack1.pop();
            if (!stack2.isEmpty()) head.val += stack2.pop();
            ListNode node = new ListNode(head.val / 10);  //node记录进位
            head.val %= 10; //head存储结果
            node.next = head; //head往前移动，指向node
            head = node;
        }
        //前导0的情况,head为node的引用，可能为0
        return head.val == 0 ? head.next : head;
    }
}