题解 | #链表相加(二)#

/**
 * 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* addInList(ListNode* head1, ListNode* head2) {
        // write code here
        if (head1 == nullptr) {
            return head2;
        }
        if (head2 == nullptr) {
            return head1;
        }
        head1 = ReverseList(head1);
        head2 = ReverseList(head2);
        ListNode* res = new ListNode(-1);
        ListNode* head = res;
        int carry = 0;
        while (head1 != NULL || head2 != NULL || carry != 0) {
            //链表不为空则取其值
            int val1 = head1 == NULL ? 0 : head1->val;
            int val2 = head2 == NULL ? 0 : head2->val;
            //相加
            int temp = val1 + val2 + carry;
            //获取进位
            carry = temp / 10;
            temp %= 10;
            //添加元素
            head->next = new ListNode(temp);
            head = head->next;
            //移动下一个
            if (head1 != NULL)
                head1 = head1->next;
            if (head2 != NULL)
                head2 = head2->next;
        }
        //结果反转回来
        return ReverseList(res->next);
    }

    ListNode* ReverseList(ListNode* pHead) {
        if (pHead == nullptr) {
            return nullptr;
        }
        ListNode* pre = nullptr;
        ListNode* cur = pHead;
        while (cur != nullptr) {
            ListNode* temp = cur -> next;
            cur -> next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }
};