我试图从链表中删除一个节点,但我仍然是双指针概念的新手,所以我尝试使用全局变量来保持头指针 . 但是,当我在删除中间节点后尝试打印列表时,得到错误的结果 .
我看到了这个问题deleting a node in the middle of a linked list,我不知道我的删除节点功能与答案有什么不同 .
这是我的代码:
#include <stdio.h>
#include <stdlib.h>
typedef unsigned char u8;
typedef struct Node node;
void addfirstnode( u8 );
void addnode( u8 );
void print( void );
void deletenode( u8 key );
void deleteonlynode();
void deletefirstnode();
struct Node
{
u8 x;
node *next;
};
node *head;
u8 length = 0;
void main( void )
{
u8 x;
printf( "\nTo add node enter 0\nTo print linked list enter 1\nTo exit press 2\nTo delete node press 3\nYour Choice:" );
scanf( "%d", &x );
if ( x == 2 )
{
printf( "\nThank You\nGood Bye" );
}
while ( x != 2 )
{
switch ( x )
{
u8 n;
u8 key;
case 0: //Add node
printf( "\nPlease enter first value:" );
scanf( "%d", &n );
if ( length == 0 )
{
addfirstnode( n );
//printf("%d",head->x);
}
else
{
addnode( n );
}
printf( "\nNode added , Thank you\n" );
break;
case 1: //Print
print();
break;
case 3: //DeleteNode
printf( "\nPlease enter value to be deleted:" );
scanf( "%d", &key );
deletenode( key );
//deletefirstnode();
break;
default:
printf( "\nInvalid Choice please try again\n" );
}
printf( "\nTo add node enter 0\nTo print linked list enter 1\nTo exit press 2\nTo delete node press 3\nYour Choice:" );
scanf( "%d", &x );
if ( x == 2 )
{
printf( "\nThank You\nGood Bye" );
}
}
//where do I use free();
}
void addfirstnode( u8 n )
{
head = ( node * ) malloc( sizeof( node ) );
head->next = NULL;
head->x = n;
length++;
}
void addnode( u8 n )
{
node *last = head;
while ( ( last->next ) != NULL )
{
last = last->next;
}
last->next = ( node * ) malloc( sizeof( node ) );
( last->next )->next = NULL;
( last->next )->x = n;
length++;
}
void print( void )
{
node *last = head;
u8 count = 1;
printf( "\n---------------------" );
if ( last == NULL )
{
printf( "\nList is empty" );
}
while ( last != NULL )
{
printf( "\nNode Number %d = %d", count, last->x );
last = last->next;
count++;
}
printf( "\n---------------------" );
printf( "\n" );
}
void deletenode( u8 key )
{
node *last = head;
//node*postlast = NULL;
if ( ( last->x == key ) && ( last->next == NULL ) )
{
deleteonlynode();
}
else
{
while ( last != NULL )
{
if ( ( last->x ) == key )
{
printf( "value to be deleted is found" );
node *temp = last->next;
last->next = last->next->next;
free( temp );
length--;
}
last = last->next;
}
}
}
void deleteonlynode()
{
printf( "\n Deleting the only node" );
free( head );
head = NULL;
length--;
}
void deletefirstnode()
{
printf( "\n Deleting the first node" );
node *temp = head;
head = head->next;
free( temp );
length--;
}
2 回答
代码正在从链表中删除错误的项目:
看到:
last
指向要删除的元素 . 但是然后代码指定temp
指向last->next
(NOTlast
),然后从列表中删除它 .因此,通过查看
node->next
而不是当前节点,它将从指针之前重新开始删除 . 基本上你的代码几乎已经存在了 .此外,我冒昧地将
last
重命名为ptr
,因为这令我感到困惑 .编辑:更新以干净地删除头部 .
您的代码似乎正在删除
last->next
,而last
应该是与该键匹配的节点 . 我猜以下代码可能会更短并删除但是,如果列表太长,则此实现(使用递归而不是循环)可能会导致StackOverFlow . 我没有测试过gcc是否会优化递归 .