题目:输入一个链表的头结点,反转该链表,并返回反转后链表的头结点。代码如下:
#include<iostream>
using namespace std;
struct Node
{
int value;
Node * link;
Node()
{
value=0;
link=NULL;
}
};
//返回反向链表的头结点
Node* Reserve(Node * head)
{
Node * pre,*next;
pre=head;
head=head->link;
pre->link=NULL;
while(head!=NULL)
{
next=head->link;
head->link=pre;
pre=head;
head=next;
}
return pre;
}
int main()
{
Node *head = new Node;
Node *tail=head;
//生成10个结点元素
for(int i=0;i<10;i++)
{
Node * newNode= new Node;
newNode->value=i+1;
tail->link=newNode;
tail=newNode;
}
//正向遍历链表
Node * interator = head;
while(interator!=NULL)
{
cout<<interator->value<<endl;
interator=interator->link;
}
//将链表反向
interator=Reserve(head);
while(interator!=NULL)
{
cout<<interator->value<<endl;
interator=interator->link;
}
//堆区内存清理
while(head!=NULL)
{
Node * next=head->link;
delete head;
head=next;
}
return 0;
}