如何在 C 中创建链表的副本到另一个地址？

Question

假设你有一个链表，你想制作不同的副本并向这个列表添加一些元素，但你不知道数量。每个链表必须有不同的地址才能更改。链表：

struct path {
   int node;
   struct path *next;
};

每次我将链表的起始地址和一个值发送给函数，函数会在另一个地址复制链表并将值添加到链表的末尾（多次）.

我实现了以下代码以在链表上创建和添加新元素：

/* Head and Tail for Linked Lists */
struct path *pathHead = NULL;
struct path *pathTail = NULL;

void newPathElement(int node) {

struct path *rv = malloc(sizeof (struct path));
rv->node = node;

if (pathHead == NULL) {
    pathHead = rv;
    pathHead -> next = NULL;
    pathTail = pathHead;
} else {
    pathTail -> next = rv;
    pathTail = rv;
    pathTail -> next = NULL;
}
}

Note I want to make different copies of a linked list. Each copy has its Head and Tail, they will be different for each copied list.

函数签名如下：

struct path *copyPath(struct path *head, struct path *tail, int node){

}

Answer 1

我们初学者应该互相帮助。:)

事实上，编写一个将新节点附加到列表的函数或将其代码合并到复制列表的函数中就足够了。

给你。

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

struct path 
{
    int node;
    struct path *next;
};

int append( struct path **head, int node )
{
    struct path *tmp = malloc( sizeof( struct path ) );
    int success = tmp != NULL;

    if ( success )
    {
        tmp->node = node;
        tmp->next = NULL;

        while ( *head ) head = &( *head )->next;

        *head = tmp;
    }

    return success;
}

struct path * copy( const struct path *source )
{
    struct path *destination = NULL;

    for ( ; source; source = source->next ) append( &destination, source->node );

    return destination;
}

void display( const struct path *head )
{
    for ( ; head; head = head->next )
    {
        printf( "%d ", head->node );
    }
}

int main(void) 
{
    struct path *first_list = NULL;
    const int N = 10;

    for ( int i = 0; i < N; i++ ) append( &first_list, i );

    display( first_list );
    putchar( '\n' );

    struct path *second_list = copy( first_list );

    display( second_list );
    putchar( '\n' );

    return 0;
}

程序输出为

0 1 2 3 4 5 6 7 8 9 
0 1 2 3 4 5 6 7 8 9

您可以通过以下方式使函数copy更有效率

struct path * copy( const struct path *source )
{
    struct path *destination = NULL;

    for ( struct path **tmp = &destination; source; source = source->next ) 
    {
        append( tmp, source->node );
        tmp = &( *tmp )->next;
    }       

    return destination;
}

如何在 C 中创建链表的副本到另一个地址？

How to create a copy of a Linked List to another address in C?

c

copy

linked-list

list

singly-linked-list