445. Add Two Numbers II

题目：

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

链接：https://leetcode.com/problems/add-two-numbers-ii/#/description

4/16/2017

61ms, 69%

其实本来想先遍历再来解决，后来发现说不能反转，哈哈那就用stack来代替好了

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
11         if (l1 == null) return l2;
12         else if (l2 == null) return l1;
13 
14         Stack<ListNode> s1 = new Stack<ListNode>();
15         Stack<ListNode> s2 = new Stack<ListNode>();
16         ListNode ret = null;
17         
18         while (l1 != null) {
19             s1.push(l1);
20             l1 = l1.next;
21         }
22         while (l2 != null) {
23             s2.push(l2);
24             l2 = l2.next;
25         }
26         int carry = 0;
27         while (!s1.empty() || !s2.empty() || carry != 0) {
28             int val = (s1.empty()? 0: s1.pop().val) + (s2.empty()? 0: s2.pop().val) + carry;
29             ListNode t = new ListNode(val % 10);
30             t.next = ret;
31             ret = t;
32             carry = val / 10;
33         }
34         return ret;
35     }
36 }

有人认为用stack是reverse链表了，我认为只要没有reverse input structure就没有问题，因为毕竟元素val和指针next都没有变化。

也有人用比较长度来算的，我自己就不想看了。

还有人认为stack不可以，但是用了recursive方法。区别也不是很大。