Question

这里我试图合并两个已排序的链接列表，但我的代码根本不起作用，所以有人能告诉我我做错了什么吗？这是执行合并两个已排序链接列表

的功能

   Node* MergeLists(Node *headA, Node* headB)

     {

     Node *temp,*ptr,*par;
      ptr=headA;
    par=headB;
    int c=0,s=0,t;
    while(ptr!=NULL)
        {
        ptr=ptr->next;
        c++;
    }
    while(par!=NULL)
        {
        par=par->next;
        s++;
    }
    t=c+s;
    //cout<<t<<"\n";
    temp=(Node*)malloc(t*sizeof(Node));
    if(headA->data <= headB->data)
        {
        temp=headA;
        ptr=headA->next;
        par=headB;
    }
    else
        {
        temp=headB;
        par=headB->next;
        ptr=headA;
    }
    while(ptr!=NULL && par!=NULL)
        {
        if(ptr->data>=par->data)
            {

            temp->next=par;
            par=par->next;
        }
        else
            {

            temp->next=ptr;
        ptr=ptr->next;
        }
        temp=temp->next;
    }
    while(ptr!=NULL)
        {
        temp->next=ptr;
        ptr=ptr->next;
        temp=temp->next;
    }
    while(par!=NULL)
        {
        temp->next=par;
        par=par->next;
        temp=temp->next;
    }
    temp->next=NULL;
     while(temp!=NULL)
        {

       cout<<temp->data<<" ";
        temp=temp->next;
    }
    return temp;
}

Answer 1

链表是基于指针的动态数据结构。

您犯的主要错误是分配一个巨大的连续堆内存，就好像您的新链接列表是数组一样。

因此，在那之后，解决方案完全错误。

下面的代码应按排序顺序合并两个已排序的链接列表，并返回指向新链接列表的指针。

Node* tempA, *tempB, *newTemp, *newHead;
tempA = headA;
tempB = headB;

newTemp = (Node*) malloc( sizeof(Node));
newHead = newTemp;

while( tempA && tempB)
{
    if( tempA->data <= tempB->data){
        newTemp->data = tempA->data;
        tempA = tempA->next;
    }
    else{
        newTemp->data = tempB->data;
        tempB = tempB->next;
    }

    newTemp->next = (Node*) malloc( sizeof(Node));
    newTemp = newTemp->next;
}

while( tempA)
{
    newTemp->data = tempA->data;
    tempA = tempA->next;

    newTemp->next = (Node*) malloc( sizeof(Node));
    newTemp = newTemp->next;
}

while( tempB)
{
    newTemp->data = tempB->data;
    tempA = tempB->next;

    newTemp->next = (Node*) malloc( sizeof(Node));
    newTemp = newTemp->next;
}

free(newTemp); //last one is useless.

return newHead;

Answer 2

我可以建议以下解决方案

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

typedef struct node
{
    int data;
    struct node *next;
} Node; 

int add( Node **head, int value )
{
    Node *tmp = malloc( sizeof( Node ) );   
    int success = tmp != NULL;

    if ( success )
    {
        tmp->data = value;

        if ( *head == NULL || value < ( *head )->data )
        {
            tmp->next = *head;
            *head = tmp;
        }
        else
        {
            while ( ( *head )->next && !( value < (*head )->next->data ) ) head = &( *head )->next;

            tmp->next = ( *head )->next;
            ( *head )->next = tmp;
        }
    }

    return success;
}

void display( Node *head )
{
    for ( ; head != NULL; head = head->next ) printf( " %d", head->data );
    printf( "\n" );
}

Node * MergeLists( Node *headA, Node *headB )
{
    Node *headC = NULL;
    Node **tailC = &headC;

    while ( headA && headB )
    {
        *tailC = malloc( sizeof( Node ) );

        if ( *tailC != NULL )
        {
            if ( headB->data < headA->data )
            {
                ( *tailC )->data = headB->data;
                headB = headB->next;
            }
            else
            {
                ( *tailC )->data = headA->data;
                headA = headA->next;
            }

            tailC = &( *tailC )->next;
        }           
    }

    while ( headA != NULL )
    {
        *tailC = malloc( sizeof( Node ) );

        if ( *tailC != NULL )
        {
            ( *tailC )->data = headA->data;
            headA= headA->next;
            tailC = &( *tailC )->next;  
        }
    }

    while ( headB != NULL )
    {
        *tailC = malloc( sizeof( Node ) );

        if ( *tailC != NULL )
        {
            ( *tailC )->data = headB->data;
            headB= headB->next;
            tailC = &( *tailC )->next;  
        }
    }

    return headC;
}

int main( void ) 
{
    const int N = 10;

    Node *headA = NULL;
    Node *headB = NULL;

    srand( ( unsigned int )time( NULL ) );

    for ( int i = 0; i < N; i++ ) add( &headA, rand() % N );
    for ( int i = 0; i < N; i++ ) add( &headB, rand() % N );

    printf( "List A:" ); display( headA );
    printf( "List B:" ); display( headB );

    Node *headC = MergeLists( headA, headB );
    printf( "List C:" ); display( headC );

    return 0;
}

程序输出可能看起来像

List A: 1 3 3 3 3 4 7 8 9 9
List B: 0 0 2 3 4 4 4 5 6 9
List C: 0 0 1 2 3 3 3 3 3 4 4 4 4 5 6 7 8 9 9 9

我试图合并两个排序的链表

2 个答案: