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