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

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        else if (l2 == null) return l1;

        Stack<ListNode> s1 = new Stack<ListNode>();
        Stack<ListNode> s2 = new Stack<ListNode>();
        ListNode ret = null;
        
        while (l1 != null) {
            s1.push(l1);
            l1 = l1.next;
        }
        while (l2 != null) {
            s2.push(l2);
            l2 = l2.next;
        }
        int carry = 0;
        while (!s1.empty() || !s2.empty() || carry != 0) {
            int val = (s1.empty()? 0: s1.pop().val) + (s2.empty()? 0: s2.pop().val) + carry;
            ListNode t = new ListNode(val % 10);
            t.next = ret;
            ret = t;
            carry = val / 10;
        }
        return ret;
    }
}

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。

更多讨论：

https://discuss.leetcode.com/category/571/add-two-numbers-ii