[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

原文：http://www.cnblogs.com/TinyBox/p/4119927.html