OJ每日一练试水第2天，聊聊链表问题

时间：2019年5月10日

1.链表合并

#include<stdio.h>
#include<stdlib.h>
typedef struct T_List  //单向链表 
{
	int data;
	T_List *pnext;
}LIST;


LIST* Read(); /* 细节在此不表 */
void Print(LIST* L); /* 细节在此不表；空链表将输出NULL */

LIST* Merge( LIST* L1, LIST* L2 );

int main()
{
	LIST *L1,*L2,*L;
	L1=Read(); 
 	L2=Read();
    L = Merge(L1, L2);
    Print(L);
    L1=0;
    L2=0;
    Print(L1);
    Print(L2);
	return 0;
	
} 
LIST* Merge( LIST* L1, LIST* L2 )
{
	LIST *pa,*pb,*pc,*pHead;	
	pa =L1;
	pb =L2;
	pHead = (LIST*)malloc(sizeof(LIST));
	
 	pc=pHead;
 	pHead=NULL;
	while(pa && pb)
	{
		if(pa->data <= pb->data)
        {
        	if(pHead==NULL)
        	{
	        	pHead=pa;
	        }
            pc->pnext = pa;
            pc = pa;
            pa = pa->pnext;
        }
        else
        {
        	if(pHead==NULL)
        	{
	        	pHead=pb;
	        }
            pc->pnext = pb;
            pc = pb;
            pb = pb->pnext;
        }
	}
	pc->pnext = pa ? pa : pb;//直接把后面的拼上 

	return pHead;
}
void Print(LIST* L)
{
	if(L==NULL) //如果链表为空 
	{
		printf("NULL\n");
		return ;
	}
	
	LIST* pTemp=(LIST*)malloc(sizeof(LIST));
	pTemp = L;
	while(pTemp!=NULL)
	{
		printf("%d ",pTemp->data);
		pTemp=pTemp->pnext;	
	}
	printf("\n");
}		
LIST* Read()
{
	int n,a;
	scanf("%d",&n);
	LIST *pHead =NULL; //链表的头 
	LIST *q =NULL; 
	while(n!=0)
	{
		LIST* pTemp=(LIST*)malloc(sizeof(LIST));
		scanf("%d",&a); 
		pTemp->data=a;
		pTemp->pnext=NULL;
		
		//将节点和链表连接(这是一个头结点不为空的链表 简称非空头链表)
		 if(pHead==NULL) //如果链表为空 
		 {
	 		pHead=pTemp;
	 
		 }else
		 {
	 		q->pnext=pTemp;
		
	 	 } 
	 	 q=pTemp;
		n--;	
	}
	return pHead;
}