혁신 공장 필기시험 문제 - 두 개의 질서정연한 체인 테이블을 하나의 새로운 체인 테이블로 통합(크기가 같아도 모든 결점을 보존)

// LinkSort.cpp :              。
//

#include "stdafx.h"
#include <iostream>
using namespace std;

struct Link
{
	int data;
	struct Link *next;
};

Link* createLink(int a[],int len)
{
	Link *head,*p,*q;
	p=new Link();
	p->data=a[0];
	head=p;
	for (int i=1;i<len;i++)
	{
		q=new Link();
		q->data=a[i];
		p->next=q;
		p=q;
	}
	p->next=NULL;
	return head;
}

void PrintLink(Link *head)
{
	Link* temp=head;
	while(temp!=NULL)
	{
		cout<<temp->data<<" ";
		temp=temp->next;
	}
	cout<<endl;
}
int getLinkLength(Link* head)
{
	int ans = 0;
	while (head!=NULL)
	{
		ans++;
		head=head->next;
	}
	return ans;
}

void Swap_Node(Link*p,Link* q)
{
	int temp=0;
	temp=p->data;
	p->data=q->data;
	q->data=temp;
}

Link* LinkSort(Link* head)
{
	Link* p,*q;
	int len=getLinkLength(head);
	for (int i=0;i<len-1;i++)
	{
		p=head;
		for (int j=0;j<len-i-1;j++)
		{
			if (p->data>p->next->data)
			{
				Swap_Node(p,p->next);
			}
			p=p->next;
		}
	}
	return head;
}
 Link * list_merge(  Link *head1,  Link *head2 )
{
	 Link *res;

	if (head1 == NULL) return head2;
	if (head2 == NULL) return head1;

	if (head1->data< head2->data)
	{
		res = head1;
		printf("res1 %d
",res->data);
		res->next = list_merge( head1->next, head2);
	} else {
		res = head2;
		printf("res2 %d
",res->data);
		res->next = list_merge( head1, head2->next);
	}

	return res;
}





int _tmain(int argc, _TCHAR* argv[])
{
	//int a[]={5,4,8,9,1,2,-1};
	int a[]={1,3,5,6,7,9};
	int b[]={2,4,6};
	int lena=sizeof(a)/sizeof(int);
	int lenb=sizeof(b)/sizeof(int);
	
	Link* aLink=createLink(a,lena);
	Link* bLink=createLink(b,lenb);
	Link*cLink = list_merge(aLink,bLink);
	
	PrintLink(cLink);

	//int len =sizeof(a)/sizeof(int);
	//Link* head=createLink(a,len);
	//PrintLink(head);
	/*head=LinkSort(head);
	PrintLink(head);*/
	system("pause");
	return 0;
}