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
思路:模拟加法。用两个指针分别指向两个链表,进行相加,用一个变量记录进位情况。如果有链表为空,则处理另一个,直到两个链表都处理完为止。
实现代码:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { ListNode preHead(0); ListNode *p = &preHead; int carry = 0; while (l1 || l2 || carry) { int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry; carry = sum / 10; p->next = new ListNode(sum % 10); p = p->next; l1 = l1 ? l1->next : l1; l2 = l2 ? l2->next : l2; } return preHead.next; } };
原文地址:http://blog.csdn.net/wolongdede/article/details/43587595