单向链表判断是否存在循环的经典方法

方法阐述: 方法(1)是建立2个指针,* 一个指针每次前进1步,另外一个指针* 每次前进2步,如果2个指针不能碰到,* 说明单向链表不存在循环;如果2个指针能够碰到,说明单向链表存在循环.这个* 方法的时间复杂性和空间复杂性是o(n)和o(1).

* **********************************/

#include <iostream>

using namespace std;

struct node

{

int value;

node* next;

};

/**链表添加节点**/

void addElement(node* head, int value)

{

node* p = head;

while ( p->next != NULL )

{

p = p->next;

}

node* q = new node;

q->next = NULL;

q->value = value;

p->next = q;

}

/**方法1 2指针追逐**/

bool isCycle(node *head)

{

node *p = head->next;

node *q = p;

while ( (p!=NULL) && (q!=NULL) && (q->next!=NULL) )

{

p = p->next;

q = q->next->next;

if ( p == q )

{

return true;

}

}

return false;

}

int main()

{

node* head = new node;

head->next = NULL;

addElement(head, 1);

addElement(head, 2);

addElement(head, 3);

addElement(head, 4);

addElement(head, 5);

addElement(head, 6);

if ( isCycle(head) == true )

{

cout << "单向链表出现循环" << endl;

}

else

{

cout << "单向链表没有循环" << endl;

}

node* p = head;

while ( p->next != NULL )

{

p = p->next;

}

p->next = head->next;

if ( isCycle(head) == true )

{