LeetCode 445. Add Two Numbers II--两个链表均按照由尾部到头部计算两个结点数值之和，保持进位

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: (6 -> 2 -> 4 -> 3) + (5 -> 9 -> 4)Output: 6 -> 8 -> 3 -> 7

import java.util.Stack;class ListNode {    int val;    ListNode next;    ListNode(int x) {        val = x;    }}public class Main {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        Stack<Integer> s1 = new Stack<Integer>();        Stack<Integer> s2 = new Stack<Integer>();        while (l1 != null) {            s1.push(l1.val);            l1 = l1.next;        }        while (l2 != null) {            s2.push(l2.val);            l2 = l2.next;        }        int sum = 0;        ListNode cur = new ListNode(0);        while (!s1.empty() || !s2.empty()) {            if (!s1.empty()) sum += s1.pop();            if (!s2.empty()) sum += s2.pop();            cur.val = sum % 10;            //为啥是“sum/10”，比如输出的[5],[5],输出应该是[1,0];如果是new ListNode(0)则输出是[0]，则错误            ListNode head = new ListNode(sum/10);//先赋值为sum/10，后面cur.val = sum % 10;会赋值            head.next = cur;//新的结点的next指针指向原有的链表头指针            cur = head;//当前指针指向头指针，表示cur指向链表头部            sum /= 10;//用来进位        }        return cur.val == 0 ? cur.next : cur;    }//addTwoNumbers    public static void main(String[] args) {//        Input: (6 -> 2 -> 4 -> 3) + (5 -> 9 -> 4)//        Output: 6 -> 8 -> 3 -> 7        ListNode A = new ListNode(6);        A.next = new ListNode(2);        A.next.next = new ListNode(4);        A.next.next.next = new ListNode(3);        ListNode B = new ListNode(5);        B.next = new ListNode(9);        B.next.next = new ListNode(4);        Main main = new Main();        ListNode C = main.addTwoNumbers(A, B);        System.out.println(C);    }}

1563 / 1563 test cases passed.
Status: Accepted
Runtime: 63 ms
Your runtime beats 46.11 % of java submissions.

T(n) = O(n)