[LeetCode] Add Two Numbers

最新推荐文章于 2024-08-12 10:14:14 发布

转载最新推荐文章于 2024-08-12 10:14:14 发布 · 61 阅读

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原文链接：http://www.cnblogs.com/zhouyoulie/p/4007504.html

本文介绍了一种解决两数相加问题的方法，通过链表形式存储数字，并以逆序方式排列各节点的数值。提供了详细的C++代码实现，展示了如何遍历两个链表并进行逐位相加，同时处理进位操作。

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

 1 class Solution {
 2 public:
 3     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
 4         if(!l1&&!l2) return NULL;
 5         if(!l1||!l2) return l1?l1:l2;
 6 
 7         int overdigit = 0;
 8         ListNode *p1 = l1;
 9         ListNode *p2 = l2;
10         ListNode *phead = NULL;
11         ListNode *pbody = NULL;
12 
13         while(p1&&p2){
14             int each_digit_val = p1->val + p2->val + overdigit;
15 
16             ListNode *pnew = new ListNode(each_digit_val%10);
17             if(!phead) phead = pnew;
18 
19             if(!pbody) pbody = pnew;
20             else{
21                 pbody->next = pnew;
22                 pbody = pnew;
23             }
24 
25             if(each_digit_val>=10) overdigit = 1;
26             else overdigit = 0;
27 
28             p1 = p1->next;p2 = p2->next;
29         }
30 
31         ListNode *p3 = p1?p1:p2;
32         while(p3||overdigit==1){
33             ListNode *pnew = new ListNode(p3?((p3->val + overdigit)%10):overdigit);
34             pbody->next = pnew;
35             pbody = pnew;
36 
37             if(((p3?(p3->val):0) + overdigit)>=10) overdigit = 1;
38             else overdigit = 0;
39             p3 = p3?p3->next:NULL;
40         }
41         return phead;
42     }
43 };

转载于:https://www.cnblogs.com/zhouyoulie/p/4007504.html