[LeetCode] Add Two Numbers

最新推荐文章于 2025-09-08 22:40:31 发布

转载最新推荐文章于 2025-09-08 22:40:31 发布 · 60 阅读

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原文链接：http://www.cnblogs.com/jcliBlogger/p/4732463.html

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#开发工具

本文详细阐述了链表相加算法的核心思想、代码实现及其时间复杂度分析，包括如何处理不同长度的链表和进位问题。

The idea is to add the two numbers (represented by linked lists) nodes after nodes and store the result in the longer list. Since we may append the additioanl carry bit after the longer list, we need to be careful not to reach the NULL pointer of the longer list.

The code is as follows. In fact, it is of time O(length(l1) + length(l2)) (the two calls oflistLength) and includes two-pass, but it is very fast in the OJ :-)

The code is as follows, about 20 lines. You may run it on the following examples and then you will understand the details of the code.

 1 class Solution {
 2 public:
 3     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
 4         if (listLength(l1) < listLength(l2))
 5             return addTwoNumbers(l2, l1);
 6         ListNode *r1 = l1, *r2 = l2;
 7         int c = 0;
 8         bool isEnd = false;
 9         while (r2) {
10             int val = r1 -> val + r2 -> val + c;
11             r1 -> val = val % 10;
12             c = val / 10;
13             if (r1 -> next) r1 = r1 -> next;
14             else isEnd = true;
15             r2 = r2 -> next;
16         }
17         while (c) {
18             int val = isEnd ? c : r1 -> val + c;
19             if (isEnd) r1 -> next = new ListNode(val % 10);
20             else r1 -> val = val % 10;
21             c = val / 10;
22             if (r1 -> next) r1 = r1 -> next;
23             else isEnd = true;
24         }
25         return l1;
26     }
27 private:
28     int listLength(ListNode* head) {
29         return head ? 1 + listLength(head -> next) : 0;
30     }
31 };