题解 | #两个链表生成相加链表# C++ 解法，链表相加（需翻转）

题解 | #两个链表生成相加链表#
C++ 解法，链表相加（需翻转）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    ListNode *reverseList(ListNode *head) {
        if (head == nullptr || head->next == nullptr) return head;
        ListNode *l = head, *r = l->next;
        l->next = nullptr;
        while (r) {
            ListNode *t = r->next;
            r->next = l;
            l = r;
            r = t;
        }
        return l;
    }
    ListNode* addInList(ListNode* head1, ListNode* head2) {
        if (head1 == nullptr) return head2;
        if (head2 == nullptr) return head1;
        ListNode *l1 = reverseList(head1);
        ListNode *l2 = reverseList(head2);
        ListNode *ans = new ListNode(0), *cur = ans;
        int carry = 0;
        while (l1 || l2 || carry) {
            int x = l1 ? l1->val : 0;
            int y = l2 ? l2->val : 0;
            int sum = x + y + carry;
            carry = sum / 10;
            sum %= 10; 
            cur->next = new ListNode(sum);
            cur = cur->next;
            if (l1) l1 = l1->next;
            if (l2) l2 = l2->next;
        }
        ans = ans->next;
        ans = reverseList(ans);
        return ans;
    }
};