请用C++写一个函数，这个函数使单链表方向逆转

是单链表方向逆转
用指针操作。。。
用while循环操作
如果用C++的类和对象表示，我再追加50分

我这个程序行不行？

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

typedef struct _Node
{
struct _Node *next;
int val;
} Node;

void reverseNodes(Node *node)
{
if (node->next == NULL)
{
return;
}
else
{
reverseNodes(node->next);
node->next->next = node;
}
}

void reverseList(Node *node)
{
reverseNodes(node);
node->next = NULL;
}

void printList(Node *node)
{
while (node)
{
printf("%d-->", node->val);
node = node->next;
}
printf("end\n");
}

int main(void)
{
Node n1, n2, n3, n4;

n1.next = &n2;
n1.val = 1;

n2.next = &n3;
n2.val = 2;

n3.next = &n4;
n3.val = 3;

n4.next = NULL;
n4.val = 4;

reverseList(&n1);

printList(&n4);

return 0;
}

你是要折腾我吧？

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

typedef struct _Node
{
struct _Node *next;
int val;
} Node;

void reverseList(Node *head)
{
Node *node = head;
int i, num = 0;
Node **array;

while (node)
{
node = node->next;
num++;
}

array = (Node **)malloc(sizeof(Node *) * num);

i = 0;
node = head;
while (node)
{
array[i] = node;
node = node->next;
i++;
}

for (i = 0; i < num; i++)
{
if (i == 0)
{
array[i]->next = NULL;
}
else
{
array[i]->next = array[i - 1];
}
}
}

void printList(Node *node)
{
while (node)
{
printf("%d-->", node->val);
node = node->next;
}
printf("end\n");
}

int main(void)
{
Node n1, n2, n3, n4;

n1.next = &n2;
n1.val = 1;

n2.next = &n3;
n2.val = 2;

n3.next = &n4;
n3.val = 3;

n4.next = NULL;
n4.val = 4;

reverseList(&n1);

printList(&n4);

return 0;
}

主函数中i.linklist(g[10],5);改为i.linklist(g,5);数组调用时用数组名

#include<iostream> using namespace std; struct LNode { int num; LNode *next; }; LNode* func(LNode *head) { LNode *pf,*pm,*pl; pf=head; if(pf) if(pm=pf->next) { while(pm) { pl=pm->next; pm->next=pf; pf=pm; pm=pl; } head->next=NULL; } return pf; } void main() { LNode *t,*head=NULL,*pos; for(int i=0;i<5;++i) { t=new LNode; cin>>t->num; if(!head){head=t;pos=head;} else {pos->next=t;pos=t;} } pos->next=NULL; pos=head; while(pos) { cout<<pos->num<<" "; pos=pos->next; } cout<<endl; head=func(head); pos=head; while(pos) { cout<<pos->num<<" "; pos=pos->next; } cout<<endl; } //////////////// 下面这个是递归函数,用这个函数替换上面的func()函数即可运行,其它地方不用改 LNode* func(LNode *head) { LNode *p; if(head) if(head->next) { p=func(head->next); head->next->next = head; head->next=NULL; return p; } else return head; }