go语言合并两个有序链表 合并两个有序的单链表,合并后依然有序

实现两个链表的合并

#include

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#include

#define OK 1

#define NULL 0

#define TURE 1

#define FLASE 0

typedef struct LNode

{

int data;

struct LNode *next;

}LNode,*LinkList;

typedef int Status;

LinkList CreateLinkList_1(void) //产生连表

{

LinkList L,p,q;

int i,Len;

printf( "\nInput the Length of the LinkList: ");

scanf( "%d ",Len);

L=(LinkList)malloc(sizeof(LNode));

L- next=NULL;

L- data=Len;

p=L;

for(i=0;i next=NULL;

printf( "Input the data: ");

scanf( "%d ",q- data);

p- next=q;

p=p- next;

}

return L;

}

Status PrintLinkList(LinkList L) //打印链表

{

LinkList q;

q=L;

if(q!=NULL)

{

printf( "The Link 's Length is:%4d\n ",q- data);

q=q- next;

printf( "L ");

do

{

printf( "-- %d ",q- data);

q=q- next;

}

while(q!=NULL);

printf( "\n ");

}

return OK;

}

LinkList LinkListSort(LinkList L) //链表排序(冒泡)

{

int i,change;

LinkList p,q,s;

for(i=L- data-1,change=TURE;i =1change;i--)

{

change=FLASE;

p=L;

q=p- next;

while(q- next!=NULL)

{

if(p- next- data q- next- data)

{

p- next=q- next;

s=q;q=p- next;

if(q- next==NULL)

{

s- next=NULL;

q- next=s;

}

else

{

s- next=q- next;

q- next=s;

}

change=TURE;

}

p=p- next;

q=q- next;

}

}

return L;

}

LinkList LinkListMerge(LinkList La,LinkList Lb) //链表的合并

{

LinkList pa,pb,pc,Lc;

Lc=(LinkList)malloc(sizeof(LNode));

Lc- data=La- data+Lb- data;

pa=La- next;

pb=Lb- next;

Lc=pc=La;

while(papb)

{

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;

free(Lb);

return Lc;

}

void main()

{

LinkList L1,L2,L3;

int n;

L1=CreateLinkList_1(); //产生链表L1 L2

PrintLinkList(L1);

L2=CreateLinkList_1();

PrintLinkList(L2);

LinkListSort(L1); //排序L1 L2

LinkListSort(L2);

printf( "\nAfter Sort:\n ");

PrintLinkList(L1); //打印排序后的链表

PrintLinkList(L2);

L3=LinkListMerge(L1,L2); //合并两个链表为一个新的有序的链表

PrintLinkList(L3);

printf( "\nPress any key to continue...\n ");

getch();

}

将两个有序单链表合并为一个有序单链表并输出其长度求大神指教

#include "stdlib.h"

#include "time.h"

struct link

{

int data;

struct link *next;

};

//不带头节点的链表

struct link *CreateLink(int n)

{ int temp,i;

struct link *head=NULL,*p=NULL,*q=NULL;

srand( n*1000);

temp = rand()%10;

head = (struct link *)malloc(sizeof(struct link));

head-data = temp;

head-next = NULL;

p = head;

for (i=1; in; i++)

{

     q = (struct link *)malloc(sizeof(struct link));

    temp = temp+rand()%8;

   q-data = temp;

   q-next = NULL;

   p-next = q;

    p = q;

}

return head;

}

void PrintLink(struct link *head)

{   int n=0;

struct link *p=head;

printf("Data: ");

while (p!=NULL)

{

   printf("%d ",p-data);

   p = p-next;

n++;

}

printf(" Nodes:%d\n",n);

}

struct link *MergeLink(struct link *h1, struct link *h2)

{

struct link *head,*tmp;

struct link *head1=h1,*head2=h2;

//哪个链表第一个节点值小,则把它的头指针作为合并后的头指针

if (h1-datah2-data)

{

   head1 = h2;

   head2 = h1;

}

head = head1;

tmp=NULL;

while (head2 != NULL)

{

   while ( (head1-next-data head2-data) head1-next!=NULL)

   {

       head1 = head1-next;

   }

   tmp = head2;

  head2 = head2-next;

   tmp-next = head1-next;

   head1-next = tmp;

}

return head;

}

int main()

{

struct link *head,*head1=NULL,*head2=NULL;

head1 = CreateLink(10);

PrintLink(head1);

head2 = CreateLink(13);

PrintLink(head2);

head = MergeLink(head1,head2);

PrintLink(head);

return 0;

}

合并两个有序列表

题目描述:

将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

示例:

输入:1-2-4, 1-3-4

输出:1-1-2-3-4-4

Go语言版本解决方案:


当前题目:go语言合并两个有序链表 合并两个有序的单链表,合并后依然有序
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