题解 | #两个链表生成相加链表#

1.翻转指针相加，啥也不是

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    ListNode* addInList(ListNode* head1, ListNode* head2) {
        // write code here
        ListNode* head1Reversed = reverse(head1);
        ListNode* head2Reversed = reverse(head2);
        int jinwei = 0;
        ListNode* res = new ListNode(-1);
        while(jinwei || head1Reversed || head2Reversed) {
            int nodeVal1 = head1Reversed?head1Reversed->val:0;
            int nodeVal2 = head2Reversed?head2Reversed->val:0;
            jinwei += (nodeVal1 + nodeVal2);
            ListNode* node = new ListNode(jinwei%10);
            node->next = res->next;
            res->next=node;
            jinwei = jinwei>=10?1:0;
            if(head1Reversed) {
                head1Reversed = head1Reversed->next;
            }
            if(head2Reversed) {
                head2Reversed = head2Reversed->next;
            }
        }
        return res->next;
    }
    
    ListNode* reverse(ListNode* head) {
        ListNode* res = new ListNode(-1);
        while(head) {
            ListNode* temp = head->next;
            head->next=res->next;
            res->next=head;
            head = temp;
        }
        return res->next;
    }
};