
难度:Easy
注意:new一个ListNode,返回的是ListNode的指针;直接调用ListNode构造函数,返回的是ListNode。
代码如下:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39
|
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { ListNode* dummy = new ListNode(-1); ListNode* cur = dummy; while(l1 != NULL && l2 != NULL) { if(l1->val < l2->val) { cur->next = l1; cur = cur->next; l1 = l1->next; } else { cur->next = l2; cur = cur->next; l2 = l2->next; } } if(l1 != NULL) { cur->next = l1; } if(l2 != NULL) { cur->next = l2; } return dummy->next; } };
|
运行结果:9ms,超过23.48%
近期评论