链表求和

2019-11-08 01:02:54

字体：大中小

来源：转载

供稿：网友

链表求和

version one

两个指针分别遍历两个链表，只需要注意好进位关系即可。其中c代表进位，temp代表链表中数字的临时加和。

ListNode *addLists(ListNode *l1, ListNode *l2) { // write your code here if (l1 == NULL) { return l2; }else if (l2 == NULL) { return l1; } int c = 0; int temp = 0; ListNode *head = new ListNode(0); ListNode *p = head; while (l1 != NULL && l2 != NULL) { temp = l1->val+l2->val + c; c = temp / 10; temp = temp % 10; p->next = new ListNode(temp); p = p->next; l1 = l1->next; l2 = l2->next; } while (l1 != NULL) { temp = l1->val + c; c = temp / 10; temp = temp % 10; p->next = new ListNode(temp); p = p->next; l1 = l1->next; } while (l2 != NULL) { temp = l2->val + c; c = temp / 10; temp = temp % 10; p->next = new ListNode(temp); p = p->next; l2 = l2->next; } if (c != 0) { p->next = new ListNode(c); } return head->next; }

上一篇：【洛谷 1601】A+B Problem（高精）

下一篇：Spring boot部署发布到linux