yxc题解

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* merge(ListNode* l1, ListNode* l2) {
        auto dummy = new ListNode(-1), tail = dummy; //这里不是tail = dummy->next;
        while (l1 && l2) {
            if (l1->val < l2->val) {
                tail = tail->next = l1;
                l1 = l1->next;
            } 
            else {
                tail = tail->next = l2;
                l2 = l2->next;
            } 
        }
        if (l1) tail->next = l1;
        if (l2) tail->next = l2;
        // tail->next = (l1 != NULL ? l1 : l2); 上面两行代码可以这样写
        return dummy->next;
    }
};