题解 | #链表相加(二)#

/**
 * 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 || !head->next) {
            return head;
        }
        ListNode *nxt = head->next;
        ListNode *res = reverseList(nxt);
        nxt->next = head;
        head->next = nullptr;
        return res;
    }

    ListNode* addInList(ListNode* head1, ListNode* head2) {
        if (!head1) return head2;
        if (!head2) return head1;
        head1 = reverseList(head1);
        head2 = reverseList(head2);
        ListNode *dummyhead = new ListNode(0), *tail = dummyhead;
        int up = 0;
        while (head1 && head2) {
            head1->val += head2->val + up;
            up = head1->val / 10;
            head1->val %= 10;
            tail->next = head1;
            tail = head1;
            head1 = head1->next;
            head2 = head2->next;
        }
        if (!head1) head1 = head2;
        while (head1) {
            head1->val += up;
            up = head1->val / 10;
            head1->val %= 10;
            tail->next = head1;
            tail = head1;
            head1 = head1->next;
        }
        if (up) {
            tail->next = new ListNode(1);
        }
        ListNode *res = dummyhead->next;
        dummyhead->next = nullptr;
        return reverseList(res);
    }
};

思路：同样是先反转链表，然后模拟加法，最后再反转回去。

因为链表与数组不同，最后注意如果有进位的话，是需要新建一个链表节点的。