标签：min http white nts dex 分析 node neu oat
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.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 | /** * 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 tmp(INT_MIN); ListNode * l3 = &tmp; while(l1 && l2){ if(l1 -> val > l2 -> val){ l3 -> next = l2; l2 = l2 -> next; } else{ l3 -> next = l1; l1 = l1 -> next; } l3 = l3 -> next; } l3 -> next = l1 == NULL? l2 : l1; return tmp.next; }}; |
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | class Solution {public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { if(l1 == NULL) return l2; if(l2 == NULL) return l1; if(l1->val < l2->val) { l1->next = mergeTwoLists(l1->next, l2); return l1; } else { l2->next = mergeTwoLists(l2->next, l1); return l2; } }}; |
标签：min http white nts dex 分析 node neu oat
原文地址：http://www.cnblogs.com/zhxshseu/p/f1a25cb59ef4d2a5b221b33668bd2ffb.html
