题解 | #链表相加(二)#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 *  ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
#include <cstddef>
class Solution {
  public:
    // 反转链表
    ListNode* inverseNodes(ListNode* head) {
        if (head == nullptr) {
            return nullptr;
        }
        auto nextNode = head->next;
        head->next = nullptr;
        while (nextNode != nullptr) {
            auto first = nextNode->next;
            nextNode->next = head;
            head = nextNode;
            nextNode = first;
        }

        return head;
    }
    /**
     * 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
        head1 = inverseNodes(head1);
        head2 = inverseNodes(head2);
        // std::cout << " nodes1: ";
        // while (head1 != nullptr) {
        //     std::cout << head1->val << " ";
        //     head1 = head1->next;
        // }
        // std::cout << " nodes2: ";
        // while (head2 != nullptr) {
        //     std::cout << head2->val << " ";
        //     head2 = head2->next;
        // }
        ListNode* head = nullptr;
        int v1 = 0, v2 = 0, v3 = 0;
        while (head1 != nullptr || head2 != nullptr) {
            v1 = 0;
            v2 = 0;
            if (head1 != nullptr) {
                v1 = head1->val;
                head1 = head1->next;
            }
            if (head2 != nullptr) {
                v2 = head2->val;
                head2 = head2->next;
            }
            v3 = v1 + v2 + v3;
            std::cout << v3 << std::endl;
            auto node = new ListNode(v3 % 10);
            node->next = head;
            v3 = v3 / 10;
            head = node;
        }
        if(v3 > 0)
        {
            auto node = new ListNode(v3);
            node->next = head;
            head = node;
        }
        return head;
    }
};