## 题目地址
https://leetcode.com/problems/add-two-numbers-ii/description/

## 题目描述

```
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

```
## 思路

由于需要从低位开始加，然后进位。 因此可以采用栈来简化操作。
依次将两个链表的值分别入栈stack1和stack2，然后相加入栈stack，进位操作用一个变量carried记录即可。

最后根据stack生成最终的链表即可。

> 也可以先将两个链表逆置，然后相加，最后将结果再次逆置。

## 关键点解析

- 栈的基本操作
- carried 变量记录进位
- 循环的终止条件设置成`stack.length > 0` 可以简化操作
- 注意特殊情况， 比如 1 + 99 = 100

## 代码
* 语言支持：JS，C++

JavaScript Code:

```js
/*
 * @lc app=leetcode id=445 lang=javascript
 *
 * [445] Add Two Numbers II
 *
 * https://leetcode.com/problems/add-two-numbers-ii/description/
 *
 * algorithms
 * Medium (49.31%)
 * Total Accepted:    83.7K
 * Total Submissions: 169K
 * Testcase Example:  '[7,2,4,3]\n[5,6,4]'
 *
 * 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.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function(l1, l2) {
    const stack1 = [];
    const stack2 = [];
    const stack = [];

    let cur1 = l1;
    let cur2 = l2;
    let curried = 0;

    while(cur1) {
        stack1.push(cur1.val);
        cur1 = cur1.next;
    }

    while(cur2) {
        stack2.push(cur2.val);
        cur2 = cur2.next;
    }

    let a = null;
    let b = null;

    while(stack1.length > 0 || stack2.length > 0) {
        a = Number(stack1.pop()) || 0;
        b = Number(stack2.pop()) || 0;

        stack.push((a + b + curried) % 10);

        if (a + b + curried >= 10) {
            curried = 1;
        } else {
            curried = 0;
        }
    }

    if (curried === 1) {
        stack.push(1);
    }

    const dummy = {};

    let current = dummy;

    while(stack.length > 0) {
        current.next = {
            val: stack.pop(),
            next: null
        }

        current = current.next
    }

    return dummy.next;
};

```
C++ Code：
```C++
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        auto carry = 0;
        auto ret = (ListNode*)nullptr;
        auto s1 = vector<int>();
        toStack(l1, s1);
        auto s2 = vector<int>();
        toStack(l2, s2);
        while (!s1.empty() || !s2.empty() || carry != 0) {
            auto v1 = 0;
            auto v2 = 0;
            if (!s1.empty()) {
                v1 = s1.back();
                s1.pop_back();
            }
            if (!s2.empty()) {
                v2 = s2.back();
                s2.pop_back();
            }
            auto v = v1 + v2 + carry;
            carry = v / 10;
            auto tmp = new ListNode(v % 10);
            tmp->next = ret;
            ret = tmp;
        }
        return ret;
    }
private:
    // 此处若返回而非传入vector，跑完所有测试用例多花8ms
    void toStack(const ListNode* l, vector<int>& ret) {
        while (l != nullptr) {
            ret.push_back(l->val);
            l = l->next;
        }
    }
};

// 逆置，相加，再逆置。跑完所有测试用例比第一种解法少花4ms
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        auto rl1 = reverseList(l1);
        auto rl2 = reverseList(l2);
        auto ret = add(rl1, rl2);
        return reverseList(ret);
    }
private:
    ListNode* reverseList(ListNode* head) {
        ListNode* prev = NULL;
        ListNode* cur = head;
        ListNode* next = NULL;
        while (cur != NULL) {
            next = cur->next;
            cur->next = prev;
            prev = cur;
            cur = next;
        }
        return prev;
    }
    
    ListNode* add(ListNode* l1, ListNode* l2) {
        ListNode* ret = nullptr;
        ListNode* cur = nullptr;
        int carry = 0;
        while (l1 != nullptr || l2 != nullptr || carry != 0) {
            carry += (l1 == nullptr ? 0 : l1->val) + (l2 == nullptr ? 0 : l2->val);
            auto temp = new ListNode(carry % 10);
            carry /= 10;
            if (ret == nullptr) {
                ret = temp;
                cur = ret;
            }
            else {
                cur->next = temp;
                cur = cur->next;
            }
            l1 = l1 == nullptr ? nullptr : l1->next;
            l2 = l2 == nullptr ? nullptr : l2->next;
        }
        return ret;
    }
};
```