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* ReverseList(ListNode* head) {
         if (head == nullptr || head->next == nullptr)
            return head;
        ListNode *cur = head; // 指向当前结点
        ListNode *pre = nullptr; // 指向反转后当前结点的下一结点
        while (cur != nullptr) {
            ListNode *temp = cur->next; // 辅助结点 -> 防止断链
            cur->next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }

    ListNode* addInList(ListNode* head1, ListNode* head2) {
        if (head1 == nullptr) return head2;
        if (head2 == nullptr) return head1;

        // 反转两个链表
        head1 = ReverseList(head1);
        head2 = ReverseList(head2);

        // 设置进位carry，开始模拟相加
        int carry = 0;
        ListNode *newHead = new ListNode(-1);
        ListNode *cur = newHead;
        while (head1 != nullptr || head2 != nullptr || carry != 0) {
            int val1 = head1 == nullptr ? 0 : head1->val;
            int val2 = head2 == nullptr ? 0 : head2->val;
            int temp = val1 + val2 + carry;
            carry = temp / 10; // 进位
            temp %= 10;
            cur->next = new ListNode(temp);
            cur = cur->next;
            if (head1 != nullptr)
                head1 = head1->next;
            if (head2 != nullptr)
                head2 = head2->next;
        }
        
        // 反转结果链表
        return ReverseList(newHead->next);
    }
};