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[leetcode] 17. Merge Two Sorted Lists
这个非常简单的题目，题目如下：
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.
就是算导里面的第一个算法讲解，但是我的C++水平实在太渣，所以对于链表的操作很蠢，但还好完成了。题解如下：
/** * 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 == NULL && l2 == NULL) { return NULL; } if (l1 == NULL || l2 == NULL) { return (l1 == NULL) ? (l2) : (l1); } ListNode *aspros, *deferos, *root; aspros = deferos = new ListNode(0); if (l1->val <= l2->val) { aspros->val = l1->val; aspros->next = NULL; l1 = l1->next; } else { aspros->val = l2->val; aspros->next = NULL; l2 = l2->next; } root = aspros; while (l1 && l2) { if (l1->val <= l2->val) { aspros = new ListNode(0); aspros->val = l1->val; aspros->next = NULL; deferos->next = aspros; deferos = aspros; l1 = l1->next; } else { aspros = new ListNode(0); aspros->val = l2->val; aspros->next = NULL; deferos->next = aspros; deferos = aspros; l2 = l2->next; } } while (l1) { aspros = new ListNode(0); aspros->val = l1->val; aspros->next = NULL; deferos->next = aspros; deferos = aspros; l1 = l1->next; } while (l2) { aspros = new ListNode(0); aspros->val = l2->val; aspros->next = NULL; deferos->next = aspros; deferos = aspros; l2 = l2->next; } return root; }};
这个就是因为我不会初始化链表，所以在一开始的地方非常蠢，作了两个判断，一个是l1与l2都为NULL的情况，另一个是它俩有一个是空的情况，可以直接弹出，接着就是合并了。
[leetcode] 17. Merge Two Sorted Lists
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