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72 lines
2.0 KiB
Markdown
72 lines
2.0 KiB
Markdown
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# LeetCode 第 2 号问题:两数相加
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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 2 号问题:两数相加。题目难度为 Medium,目前通过率为 33.9% 。
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### 题目描述
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给出两个 **非空** 的链表用来表示两个非负的整数。其中,它们各自的位数是按照 **逆序** 的方式存储的,并且它们的每个节点只能存储 **一位** 数字。
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如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。
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您可以假设除了数字 0 之外,这两个数都不会以 0 开头。
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**示例:**
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```
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输入:(2 -> 4 -> 3) + (5 -> 6 -> 4)
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输出:7 -> 0 -> 8
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原因:342 + 465 = 807
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```
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### 题目解析
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设立一个表示进位的变量`carried`,建立一个新链表,把输入的两个链表从头往后同时处理,每两个相加,将结果加上`carried`后的值作为一个新节点到新链表后面。
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### 动画描述
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![](https://bucket-1257126549.cos.ap-guangzhou.myqcloud.com/20181117122234.gif)
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### 代码实现
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```
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/// 时间复杂度: O(n)
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/// 空间复杂度: O(n)
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class Solution {
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public:
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ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
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ListNode *p1 = l1, *p2 = l2;
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ListNode *dummyHead = new ListNode(-1);
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ListNode* cur = dummyHead;
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int carried = 0;
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while(p1 || p2 ){
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int a = p1 ? p1->val : 0;
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int b = p2 ? p2->val : 0;
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cur->next = new ListNode((a + b + carried) % 10);
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carried = (a + b + carried) / 10;
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cur = cur->next;
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p1 = p1 ? p1->next : NULL;
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p2 = p2 ? p2->next : NULL;
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}
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cur->next = carried ? new ListNode(1) : NULL;
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ListNode* ret = dummyHead->next;
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delete dummyHead;
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return ret;
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}
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};
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```
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![](https://bucket-1257126549.cos.ap-guangzhou.myqcloud.com/blog/fz0rq.png)
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