题解 | #链表相加(二)#参考答案思路解题
链表相加(二)
https://www.nowcoder.com/practice/c56f6c70fb3f4849bc56e33ff2a50b6b
/** * struct ListNode { * int val; * struct ListNode *next; * ListNode(int x) : val(x), next(nullptr) {} * }; */ #include <bits/types/struct_tm.h> #include <cstddef> class Solution { public: /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param head1 ListNode类 * @param head2 ListNode类 * @return ListNode类 */ ListNode* reverse(ListNode* head) { ListNode* newhead = new ListNode(0); ListNode* pre = NULL; newhead->next = head; while (head) { ListNode* tmp = head->next; newhead->next = head; head->next = pre; pre = head; head = tmp; } return newhead->next; } ListNode* addInList(ListNode* head1, ListNode* head2) { if (!head1) return head2; if (!head2) return head1; ListNode* addhead = new ListNode(0); ListNode* begin = addhead; head1 = reverse(head1); head2 = reverse(head2); int innumber = 0; while (head1 && head2) { int val = head1->val + head2->val + innumber; innumber = val / 10; ListNode* addnode = new ListNode(val % 10); begin->next = addnode; begin = begin->next; head1 = head1->next; head2 = head2->next; } if (!head1 && !head2) { return reverse(addhead->next); } if (!head1) { while (head2) { int val = head2->val + innumber; innumber = val / 10; ListNode* addnode = new ListNode(val % 10); begin->next = addnode; begin = begin->next; head2 = head2->next; } if (innumber == 1) { ListNode* addnode = new ListNode(innumber); begin->next = addnode; } } if (!head2) { while (head1) { int val = head1->val + innumber; innumber = val / 10; ListNode* addnode = new ListNode(val % 10); begin->next = addnode; begin = begin->next; head1 = head1->next; } if (innumber == 1) { ListNode* addnode = new ListNode(innumber); begin->next = addnode; } } return reverse(addhead->next); // write code here } };