Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
题目的大概意思是:将两个有序单链表合并成一个有序链表。
这道题难度等级:简单
我的思路是:固定一个链表不动,将其中另一个链表中的数插到固定链表中的合适位置,具体细节酌情处理。
代码如下:
/** * 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) { if (!l1){return l2;} if (!l2){return l1;} ListNode* p, * q, *s, *r; (l1->val > l2->val) ? (p = l2,q = l1) : (p = l1, q = l2);//p小 q大 p不动 s = p; while(p->next && q){ (q->val >= p->next->val) ? (p = p->next) : (r = q->next, q->next = p->next, p->next = q, q = r, p = p->next); } if (!p->next){p->next = q;} return s; } };提交代码,顺利AC掉,Runtime:8ms。