题解 | #链表相加(二)#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /*
        反转链表
    */
    ListNode* reverseList(ListNode *head) {
        if (head == nullptr || head->next == nullptr) {
            return head;
        }
        // 头插法
        ListNode tmp(0);
        ListNode *newHead = &tmp, *second = head->next;
        while (head) {
            head->next = newHead->next;
            newHead->next = head;
            head = second;
            second = head->next;
        }
        return newHead->next;
    }
    /**
     * 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) {
        head1 = reverseList(head1);
        head2 = reverseList(head2);
        ListNode res(0);
        ListNode *cur = &res;
        int c = 0; // 进位
        while (head1 && head2) {
            int tmp = head1->val + head2->val + c;
            c = tmp / 10;
            ListNode *newNode = new ListNode(tmp % 10);
            newNode->next = cur->next;
            cur->next = newNode;
            head1 = head1->next;
            head2 = head2->next;
        }
        while (head1) {
            int tmp = head1->val + c;
            c = tmp / 10;
            ListNode *newNode = new ListNode(tmp % 10);
            newNode->next = cur->next;
            cur->next = newNode;
            head1 = head1->next;
        }
        while (head2) {
            int tmp = head2->val + c;
            c = tmp / 10;
            ListNode *newNode = new ListNode(tmp % 10);
            newNode->next = cur->next;
            cur->next = newNode;
            head2 = head2->next;
        }
        if (c != 0) {
            ListNode *newNode = new ListNode(c);
            newNode->next = cur->next;
            cur->next = newNode;
        }
        return cur->next;
    }
};