两数相加||

给你两个 非空 链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。 你可以假设除了数字 0 之外，这两个数字都不会以零开头。

进阶： 如果输入链表不能修改该如何处理？换句话说，你不能对列表中的节点进行翻转。

示例： 输入：(7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4) 输出：7 -> 8 -> 0 -> 7

一种很烂的方法，将l1,l2翻转，每个结点加起来，得到的结果相翻转再返回。

struct ListNode* reverse(struct ListNode* head) { if (head == NULL || head->next == NULL) return head; struct ListNode* before = NULL; struct ListNode* p = head; struct ListNode* later = p->next; while (p) { later = p->next; p->next = before; before = p; p = later; } return before; } struct ListNode *addTwoNumbers(struct ListNode *l1, struct ListNode *l2) { struct ListNode *h1 = reverse(l1); struct ListNode *h2 = reverse(l2); struct ListNode *head = (struct ListNode*)malloc(sizeof(struct ListNode)); head->next = NULL; struct ListNode *record = head; int flag = 0; while (h1 && h2) { struct ListNode *p = (struct ListNode*)malloc(sizeof(struct ListNode)); p->val = h1->val + h2->val + flag; flag = p->val / 10; p->val %= 10; p->next = NULL; record->next = p; record = p; h1 = h1->next; h2 = h2->next; } while (h1) { h1->val += flag; flag = h1->val / 10; h1->val %= 10; record->next = h1; record = h1; h1 = h1->next; } while (h2) { h2->val += flag; flag = h2->val / 10; h2->val %= 10; record->next = h2; record = h2; h2 = h2->next; } if (flag) { head->val = 1; record->next = head; record = head->next; head->next = NULL; l1 = reverse(record); } else l1 = reverse(head->next); return l1; }

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/add-two-numbers-ii 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。