题解 | #链表相加(二)#
链表相加(二)
https://www.nowcoder.com/practice/c56f6c70fb3f4849bc56e33ff2a50b6b
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverse(ListNode* pHead) {
ListNode* lst = nullptr;
while (pHead) {
ListNode* tmp = pHead -> next;
pHead -> next = lst;
lst = pHead;
pHead = tmp;
}
return lst;
}
ListNode* addInList(ListNode* head1, ListNode* head2) {
if (head1 == nullptr) return head2;
if (head2 == nullptr) return head1;
head1 = reverse(head1);
head2 = reverse(head2);
int cy = 0;
int c1, c2, tmp;
ListNode* lst = new ListNode(1);//将头节点设为1，如果最后进位为1就返回头节点，进位为0就返回第二个节点，不要头节点
while (head1 || head2) {
if (head1 && head2) {
c1 = head1 -> val;
c2 = head2 -> val;
tmp = c1 + c2 + cy;
} else if (head1) {
c1 = head1 -> val;
tmp = c1 + cy;
} else if (head2) {
c2 = head2 -> val;
tmp = c2 + cy;
}
cy = tmp / 10;
tmp = tmp % 10;
ListNode* node = new ListNode(tmp);
node -> next = lst -> next;//头插法，生成链表刚好是倒序
lst -> next = node;
if (head1) head1 = head1 -> next;
if (head2) head2 = head2 -> next;
}
if(cy)
return lst;
return lst -> next;
}
};
