You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.
Example
Given 7->1->6 + 5->9->2. That is, 617 + 295.
Return 2->1->9. That is 912.
Given 3->1->5 and 5->9->2, return 8->0->8.
LeetCode上的原题，请参见我之前的博客Add Two Numbers。
class Solution { public: /** * @param l1: the first list * @param l2: the second list * @return: the sum list of l1 and l2 */ ListNode *addLists(ListNode *l1, ListNode *l2) { ListNode *dummy = new ListNode(-1), *cur = dummy; int carry = 0; while (l1 || l2) { int num1 = l1 ? l1->val : 0; int num2 = l2 ? l2->val : 0; int sum = num1 + num2 + carry; cur->next = new ListNode(sum % 10); carry = sum / 10; if (l1) l1 = l1->next; if (l2) l2 = l2->next; cur = cur->next; } if (carry) cur->next = new ListNode(1); return dummy->next; } };
本文转自博客园Grandyang的博客，原文链接：两个数字相加[LintCode] Add Two Numbers ，如需转载请自行联系原博主。
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