LeetCode Merge Two Sorted Lists

Merge Two Sorted Lists

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.

入门题目了，严蔚敏的书本好像一开始就介绍了这样的题目，以前死背了好多次，始终背不出来，现在已经不用背了，感觉随时都能写出不比书本差的程序。

本题就是考链表的操作了。

链表的操作是要练出感觉来的。

下面的程序提供到位操作，把链表2直接插入链表1中。

/**
 * 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 (!l2) return l1;
	    if (!l1) return l2;
		
		ListNode *h = l1;

		if (l1->val > l2->val)
		{
			ListNode *t = l2->next;
			l2->next = l1;
			h = l1 = l2;
			l2 = t;
		}

		while (l1->next && l2)
		{
			if (l1->next->val > l2->val)
			{
				ListNode *t = l2->next;
				l2->next = l1->next;
				l1->next = l2;
				l2 = t;
			}
			l1 = l1->next;
		}

		if (l2) l1->next = l2;

		return h;
	}
};

//2014-2-8 update
	ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) 
	{
		ListNode dummy(0);
		ListNode *p = &dummy;

		while (l1 && l2)
		{
			if (l1->val > l2->val)
			{
				p->next = l2;
				l2 = l2->next;
			}
			else
			{
				p->next = l1;
				l1 = l1->next;
			}
			p = p->next;
		}
		if (l1) p->next = l1;
		else if (l2) p->next = l2;

		return dummy.next;
	}