题解 | #链表相加(二)#

class Solution {
  public:
    ListNode* addInList(ListNode* head1, ListNode* head2) {
        ListNode* l1 = reverseList(head1);//翻转链表1
        ListNode* l2 = reverseList(head2);//翻转链表2

        ListNode* head = nullptr;
        ListNode* tail = nullptr;
        int carry = 0;					//进位
        while (l1 || l2) {
            int n1 = l1 ? l1->val : 0;
            int n2 = l2 ? l2->val : 0;
            int sum = n1 + n2 + carry;
            if (!head) {
                head = new ListNode(sum % 10);
                tail = head;
            } else {
                tail->next = new ListNode(sum % 10);
                tail = tail->next;
            }
            if (l1) l1 = l1->next;//循环
            if (l2) l2 = l2->next;
            carry = sum / 10;
        }
        tail->next = carry ? new ListNode(carry) : nullptr;//尾部节点
        ListNode* pHead = reverseList(head);
        return pHead;
    }

    ListNode* reverseList(ListNode* head) {//翻转链表
        ListNode* cur = head;
        ListNode* pre = nullptr;
        while (cur) {
            ListNode* temp = cur->next;
            cur->next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }
};