LeetCode2两数相加-C语言
给出两个?非空?的链表用来表示两个非负的整数。其中,它们各自的位数是按照?逆序?的方式存储的,并且它们的每个节点只能存储?一位?数字。 如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。 您可以假设除了数字 0 之外,这两个数都不会以 0?开头。 示例: 输入:(2 -> 4 -> 3) + (5 -> 6 -> 4) 输出:7 -> 0 -> 8 原因:342 + 465 = 807 /** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* addTwoNumbers(struct ListNode* l1,struct ListNode* l2) { struct ListNode *l3 = (struct ListNode *)malloc(sizeof(struct ListNode)); struct ListNode *p1 = l1,*p2 = l2,*p3 = l3; p3 -> next = NULL; p3 -> val = 0; while(p1 != NULL && p2 != NULL){ struct ListNode *node = (struct ListNode *)malloc(sizeof(struct ListNode)); node -> val = 0; node -> next = NULL; p3 -> next = node; p3 -> val += p1 -> val + p2 -> val; if(p3 -> val >= 10){ p3 -> val -= 10; node -> val = 1; } p1 = p1 -> next; p2 = p2 -> next; if(p1 == NULL && p2 == NULL && p3 -> next -> val == 0) {p3 -> next = NULL;} p3 = p3 -> next; } while(p1 != NULL){ struct ListNode *node = (struct ListNode *)malloc(sizeof(struct ListNode)); node -> val = 0; node -> next = NULL; p3 -> next = node; p3 -> val += p1 -> val; if(p3 -> val >= 10){ p3 -> val -= 10; node -> val = 1; } p1 = p1 -> next; if(p1 == NULL && p3 -> next -> val == 0) p3 -> next = NULL; p3 = p3 -> next; } while(p2 != NULL){ struct ListNode *node = (struct ListNode *)malloc(sizeof(struct ListNode)); node -> val = 0; node -> next = NULL; p3 -> next = node; p3 -> val += p2 -> val; if(p3 -> val >= 10){ p3 -> val -= 10; node -> val = 1; } p2 = p2 -> next; if(p2 == NULL && p3 -> next -> val == 0) p3 -> next = NULL; p3 = p3 -> next; } return l3; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |