2019-05-02 15:59:01 +08:00
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# LeetCode 第 445 号问题：两数相加 II
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> 本文首发于公众号「五分钟学算法」，是[图解 LeetCode ](<https://github.com/MisterBooo/LeetCodeAnimation>)系列文章之一。
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>
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> 个人网站：[https://www.cxyxiaowu.com](https://www.cxyxiaowu.com)
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题目来源于 LeetCode 上第 445 号问题：两数相加 II。题目难度为 Medium，目前通过率为 48.8% 。
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### 题目描述
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给定两个**非空**链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储单个数字。将这两数相加会返回一个新的链表。
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你可以假设除了数字 0 之外，这两个数字都不会以零开头。
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**进阶:**
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如果输入链表不能修改该如何处理？换句话说，你不能对列表中的节点进行翻转。
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**示例:**
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```
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输入: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
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输出: 7 -> 8 -> 0 -> 7
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```
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### 题目解析
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由于计算时要保证最右边的数对齐，那么很自然的想到先用**栈**存放链表中的每个值，然后依次计算。由于相加时可能产生进位，所以使用一个flag表示是否有进位。
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提示：若栈中元素相加结束之后仍有进位，则需要新加入一个头结点。
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### 动画描述
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### 代码实现
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```python
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class Solution:
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def addTwoNumbers(self, l1, l2):
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# 分别入栈
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stack1 = []
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stack2 = []
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while l1:
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stack1.append(l1.val)
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l1 = l1.next
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while l2:
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stack2.append(l2.val)
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l2 = l2.next
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flag = 0
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head = None
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while stack1 or stack2 or flag != 0:
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if stack1:
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flag += stack1.pop()
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if stack2:
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flag += stack2.pop()
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node = ListNode(flag % 10)
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node.next = head
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head = node
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flag = flag // 10
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return head
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```
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2019-11-14 11:00:28 +08:00
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