创新工场笔试题目--将两个有序的链表合并为一个新的链表（保留所有结点，即使大小相同）

直接上代码：

// LinkSort.cpp : 定义控制台应用程序的入口点。
//

#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\n",res->data);
  res->next = list_merge( head1->next, head2);
 } else {
  res = head2;
  printf("res2 %d\n",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;
}

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

// LinkSort.cpp : 定义控制台应用程序的入口点。//#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\n",res->data);res->next = list_merge( head1->next, head2);} else {res = head2;printf("res2 %d\n",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;}