问题描述

C语言中单向的链表反转

node* reverse1(node * head)
{
node*p,*q,*r;
p = head;

q = head->next;

head->next = NULL;
while(q){

r = q->next;

q->next = p;

p = q;

q = r;
}

head=p;
/* while(p)
{
printf("%d ",p->num);
p=p->next;
}*/
printf("
");
return head;

}

解决方案

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

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

node*
reverse1(node * head)
{
        node*p,*q,*r;

        if( head != NULL )
        {
                p = head;
                q = head->next;

                head->next = NULL;
                while(q){
                        r = q->next;
                        q->next = p;

                        p = q;
                        q = r;
                }

                head=p;
        }
        return head;
}

int
main( void )
{
        node *link = NULL;
        node *tmp = NULL;
        int i = 0;

        for( i = 0; i < 5; i++ )
        {
                tmp = malloc( sizeof( node ) );
                if( tmp == NULL )
                        return -1;

                tmp->data = i;
                tmp->next = link;
                link = tmp;
        }

        tmp = link;
        printf( "before reversal:
" );
        while( tmp )
        {
                printf( "%d ", tmp->data );
                tmp = tmp->next;
        }
        putchar( '
' );

        link = reverse1( link );
        printf( "after reversal:
" );
        tmp = link;
        while( tmp )
        {
                printf( "%d ", tmp->data );
                tmp = tmp->next;
        }
        putchar( '
' );

        return 0;
}

测了一下，应该可以使吧

解决方案二：

示意图

代码实现

[cpp]?view
plaincopy

#include<stdio.h>??
???
struct?ListNode{??
????int&nb......
答案就在这里：C 单向链表反转
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