7-18 merging of two ordered linked list sequences (20 minutes)

Given two non descending list sequences S1 and S2, the design function constructs a new non descending list S3 after the combination of S1 and S2.

Input format:

The input is divided into two lines. Each line gives a non descending sequence composed of several positive integers. The end of the sequence is represented by − 1 (− 1 does not belong to this sequence). Numbers are separated by spaces.

Output format:

Output a new non descending linked list after merging in a row. Separate the numbers with spaces. No extra spaces are allowed at the end. If the new linked list is empty, output NULL.

Input example:

1 3 5 -1
2 4 6 8 10 -1

Output example:

1 2 3 4 5 6 8 10

Method 1: array representation

#include<stdio.h>
#define N 500000
int main()
{
	int a[N],b[N],c[N],asize=0,bsize=0,csize=0;
	int temp,i=0,j=0,k;
	while(scanf("%d",&temp)!=EOF&&temp!=-1)
	{
		a[asize++]=temp;
	}
	while(scanf("%d",&temp)!=EOF&&temp!=-1)
	{
		b[bsize++]=temp;
	}
	while(i<asize&&j<bsize)
	{
		if(a[i]<b[j])
		{
			c[csize++]=a[i];
			i++;
		}
		else
		{
			c[csize++]=b[j];
			j++;
		}
	}
	if(i==asize)
	{
		for(k=j;k<bsize;k++)
		{
			c[csize++]=b[k];
		}
	}
	else
	{
		for(k=i;k<asize;k++)
		{
			c[csize++]=a[k];
		}
	}
	if(csize>0)
	{
		printf("%d",c[0]);
		for(i=1;i<csize;i++)
		{
			printf(" %d",c[i]);
		}
	}
	else printf("NULL");
}

Advantages: simple and easy to understand, fast coding

Disadvantage: there will be a segment error in a big data test, only 16 points

Method 2: chain table representation

#include<stdio.h>
#include<stdlib.h>
typedef struct Node *LNode;
typedef struct Node
{
	int data;
	LNode Next;
}Node;
void Insert(LNode head)
{
	LNode p,q=head;
	int temp;
	while(scanf("%d",&temp)!=EOF&&temp!=-1)
	{
		p=(LNode)malloc(sizeof(Node));
		p->data=temp;
		p->Next=NULL;
		q->Next=p;
		q=p;
	}
}
void cmp(LNode head1,LNode head2,LNode head3)
{
	LNode p1,p2,p3,q3;
	p1=head1->Next;
	p2=head2->Next;
	p3=q3=head3;
	while(p1&&p2)
	{
		if(p1->data<p2->data)
		{
			p3=(LNode)malloc(sizeof(Node));
			p3->data=p1->data;
			p3->Next=NULL;
			
			q3->Next=p3;
			q3=p3;
			p1=p1->Next;
		}
		else
		{
			p3=(LNode)malloc(sizeof(Node));
			p3->data=p2->data;
			p3->Next=NULL;
			
			q3->Next=p3;
			q3=p3;
			p2=p2->Next;
		}
	}
	
	if(p1)
	{
		while(p1)
		{
			p3=(LNode)malloc(sizeof(Node));
			p3->data=p1->data;
			p3->Next=NULL;
			
			q3->Next=p3;
			q3=p3;
			p1=p1->Next;
		}
	}
	else
	{
		while(p2)
		{
			p3=(LNode)malloc(sizeof(Node));
			p3->data=p2->data;
			p3->Next=NULL;
			
			q3->Next=p3;
			q3=p3;
			p2=p2->Next;
		}
	}
}
void Print(LNode head3)
{
	LNode p3;
	p3=head3->Next;
	if(!p3) printf("NULL");
	else 
	{
		printf("%d",p3->data);
		p3=p3->Next;
	}
	while(p3)
	{
		printf(" %d",p3->data);
		p3=p3->Next;
	}
}
int main()
{
	LNode head1,head2,head3;
	head1=(LNode)malloc(sizeof(Node));
	head2=(LNode)malloc(sizeof(Node));
	head3=(LNode)malloc(sizeof(Node));
	
	head1->Next=NULL;head2->Next=NULL;head3->Next=NULL;
	
	Insert(head1);   //Used to set up linked list 1 
	Insert(head2);   //Used to create a list 2 
	cmp(head1,head2,head3); //Compare list 1 with list 2, and fill in the comparison results in list 3 
	Print(head3);   //Output list 3 
	
}

Advantages: all test points pass

Disadvantages: too complex, slow encoding speed