题解 | #链表相加(二)#
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) {} * }; */ 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){ if(!head) return head; ListNode *pre=NULL; ListNode *cur=head; while(cur){ ListNode *tmp=cur->next; cur->next=pre; pre=cur; cur=tmp; } return pre; } ListNode* addInList(ListNode* head1, ListNode* head2) { // write code here if(!head1) return head2; if(!head2) return head1; head1=reverse(head1); head2=reverse(head2); ListNode* head = new ListNode(0); ListNode* cur = head; int uper = 0; while(head1 || head2){ int val=uper; if(head1) { val+=head1->val; head1=head1->next; } if(head2) { val+=head2->val; head2=head2->next; } uper=val/10; cur->next=new ListNode(val%10); cur=cur->next; } if(uper!=0){ cur->next=new ListNode(uper); } return reverse(head->next); } };