Add Two Numbers
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
链表处理
class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ListNode* result_head = new ListNode(0); ListNode* p; p = result_head; while (l1 != NULL&&l2 != NULL) { ListNode* tmp = new ListNode(0); (*p).val += (*l1).val + (*l2).val; (*p).next = tmp; if ((*p).val >= 10) { (*p).val %= 10; (*tmp).val += 1; } p = tmp; l1 = (*l1).next; l2 = (*l2).next; } if (l1 == NULL&&l2 == NULL) {} else { if (l1 == NULL&&l2 != NULL) { while (l2 != NULL) { (*p).val += (*l2).val; ListNode* tmp = new ListNode(0); (*p).next = tmp; if ((*p).val >= 10) { (*p).val %= 10; (*tmp).val += 1; } p = tmp; l2 = (*l2).next; } } else { if (l1 != NULL&&l2 == NULL) { while (l1 != NULL) { (*p).val += (*l1).val; ListNode* tmp = new ListNode(0); (*p).next = tmp; if ((*p).val >= 10) { (*p).val %= 10; (*tmp).val += 1; } p = tmp; l1 = (*l1).next; } } } } p = result_head; while (((*p).next)->next != NULL) p = (*p).next; if (((*p).next)->val == 0) { delete (*p).next; (*p).next = NULL; } return result_head; } };
原文：http://www.cnblogs.com/chenwanqq-leetcode/p/4596347.html