合并两个有序链表

题目描述：

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。
(hint: 请务必使用链表。)

输入：

输入可能包含多个测试样例，输入以EOF结束。
对于每个测试案例，输入的第一行为两个整数n和m(0<=n<=1000, 0<=m<=1000)：n代表将要输入的第一个链表的元素的个数,m代表将要输入的第二个链表的元素的个数。
下面一行包括n个数t(1<=t<=1000000)：代表链表一中的元素。接下来一行包含m个元素，s(1<=t<=1000000)。

输出：

对应每个测试案例，
若有结果，输出相应的链表。否则，输出NULL。

样例输入：

5 2
1 3 5 7 9
2 4
0 0

样例输出：

1 2 3 4 5 7 9
NULL

----------------------------------------------------

/*****************************
有序链表合并
by Rowandjj
2014/7/16
*******************************/
#include<iostream>
#include<stdlib.h>
#include<stdio.h>
using namespace std;
typedef struct Node
{
	int data;
	struct Node *next;
}Node,*List;
Node* MergeList(List la,List lb)
{
	List lc;
	if(!la)
	{
		return lb;	
	}else if(!lb)
	{
		return la;
	}
	Node *pa = la,*pb = lb,*pc;
	if(pa->data < pb->data)
	{
		pc = pa;
		pa = pa->next;
	}else
	{
		pc = pb;
		pb = pb->next;
	}
	lc = pc;
	while(pa && pb)
	{
		if(pa->data < pb->data)
		{
			pc->next = pa;
			pc = pa;
			pa = pa->next;
		}else
		{
			pc->next = pb;
			pc = pb;
			pb = pb->next;
		}
	}
	pc->next = pa ? pa : pb;
	return lc;
}
int main()
{
	int n,m,i;
	int data;
	while(scanf("%d %d",&n,&m) != EOF)
	{	
		Node *pCur;
		List nL = NULL;
		if(n>0)
		{
			nL = (Node*)malloc(sizeof(Node));
			if(!nL)
			{
				exit(0);
			}
			cin>>data;
			nL->data = data;
			nL->next = NULL;
			
			pCur = nL;
			for(i = 0; i < n-1; i++)
			{
				Node *newNode = (Node*)malloc(sizeof(Node));
				if(!newNode)
				{
					exit(-1);
				}
				cin>>data;
				newNode->data = data;
				newNode->next = NULL;
				pCur->next = newNode;
				pCur = newNode;
			}
		}
		List mL = NULL;
		if(m>0)
		{
			mL = (List)malloc(sizeof(Node));
			if(!mL)
			{
				exit(0);
			}
			cin>>data;
			mL->data = data;
			mL->next = NULL;
			pCur = mL;
			for(i = 0; i < m-1; i++)
			{
				Node *nodeNew = (Node*)malloc(sizeof(Node));
				if(!nodeNew)
				{
					exit(-1);
				}
				cin>>data;
				nodeNew->data = data;
				nodeNew->next = NULL;
				pCur->next = nodeNew;
				pCur = nodeNew;
			}
		}
		
		List newList = MergeList(nL,mL);
		if(newList != NULL)
		{
			Node *p = newList;
			while(p)
			{
				if(p->next == NULL)
				{
					cout<<p->data<<endl;
				}else
				{
					cout<<p->data<<" ";
				}
				
				p = p->next;
			}
		}else
		{
			cout<<"NULL"<<endl;	
		}
	}
	return 0;
}

//递归版本
Node* MergeList_2(List la,List lb)
{
	if(!la)
	{
		return lb;
	}else if(!lb)
	{
		return la;
	}
	Node *phead = NULL;
	if(la->data < lb->data)
	{
		phead = la;
		phead->next = MergeList_2(la->next,lb);
	}else
	{
		phead = lb;
		phead->next = MergeList_2(la,lb->next);
	}
	return phead;
}