[LeetCode #2] Add Two Numbers

weixin_30799995

于 2016-09-24 09:16:00 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/amadis/p/5902442.html

本文介绍了一个使用链表表示非负整数并实现两个链表相加的方法。链表中的数字以逆序方式存储，每个节点包含一个数字。通过迭代两个链表节点，逐位进行加法运算，并将进位值传递到下一位。

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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         ListNode dummy(0);
 5         ListNode *p = &dummy;
 6         ListNode *p1 = l1, *p2 = l2;
 7         
 8         int carry = 0;
 9         
10         while (p1 || p2){
11             int n1 = p1 ? p1->val : 0;
12             int n2 = p2 ? p2->val : 0;
13             int sum = n1 + n2 + carry;
14             carry = sum / 10;
15             p->next = new ListNode(sum % 10);
16             p = p->next;
17             if (p1) p1 = p1->next;
18             if (p2) p2 = p2->next;
19         }
20         
21         if (carry){
22             p->next = new ListNode(carry);
23         }
24         
25         return dummy.next;
26     }
27 };