/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        //利用递归
        //哪个list为null，则说明另一个list存了所有的结果
        if (l1 == NULL) return l2;
        if (l2 == NULL) return l1;
        //小的数放在最前面
        if (l1->val >= l2 -> val) l2->next = mergeTwoLists(l1, l2->next);
        else {
            l1->next = mergeTwoLists(l1->next, l2);
            //确保最终结果在l2上
            l2 = l1;
        }
        
        return l2;
        
    }
};