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* addInList(ListNode* head1, ListNode* head2) {
        // write code here
	//反转两个链表，使链表表头是低位
        ListNode* p1 = nullptr;
        ListNode* p2 = head1;
        ListNode* p3 = head1->next;
        while (p3) {
            p2->next = p1;
            p1 = p2;
            p2 = p3;
            p3 = p3->next;
        }
        p2->next = p1;
        head1 = p2;

        p1 = nullptr;
        p2 = head2;
        p3 = head2->next;
        while (p3) {
            p2->next = p1;
            p1 = p2;
            p2 = p3;
            p3 = p3->next;
        }
        p2->next = p1;
        head2 = p2;

	//添加一个空的头节点，指向两数相加结果的低位
        ListNode* pre = new ListNode(0);
        ListNode* cur = pre;
        int x = 0, y = 0, sum = 0, carry = 0;
        while (head1 || head2) {
            x = head1 == nullptr ? 0 : head1->val;
            y = head2 == nullptr ? 0 : head2->val;
            sum = x + y + carry;
            carry = sum / 10;
            sum = sum % 10;

            cur->next = new ListNode(sum);
            cur = cur->next;

            if (head1) head1 = head1->next;
            if (head2) head2 = head2->next;
        }
        if (carry == 1) {
            cur->next = new ListNode(1);
        }
	//反转结果链表
        p1 = nullptr;
        p2 = pre->next;
        p3 = p2->next;
        while (p3) {
            p2->next = p1;
            p1 = p2;
            p2 = p3;
            p3 = p3->next;
        }
        p2->next = p1;
        return p2;

    }
};