判断单链表是否带环

Posted 2022-12-07 ssopp24

//这个题思路不难，我们只要定义两个指针，一快一慢，快指针一次走两步，慢指针一次走一步
//如果它们相遇，则单链表构成了环形结构。 如果未相遇， 而快指针走到了链表结尾，则单链表未构成环形结构
#include <iostream>
using namespace std;


struct ListNode
	int m_nKey;
	ListNode* m_pNext;

	ListNode( ListNode* pNext, int value )
		: m_pNext( pNext )
		, m_nKey( value )
	
;


bool IsRing( ListNode* pListHead );
void push_back( ListNode** pHead, int value );
ListNode* Find( ListNode* pListHead, int value );


int main( )

	ListNode* pHead = NULL;
	//push_back( &pHead, 1 );
	//push_back( &pHead, 2 );
	//push_back( &pHead, 3 );
	//push_back( &pHead, 4 );
	//push_back( &pHead, 5 );	
	
	push_back( &pHead, 1 );
	push_back( &pHead, 2 );
	push_back( &pHead, 3 );
	push_back( &pHead, 4 );
	push_back( &pHead, 5 );

	ListNode* pNode3 = Find( pHead, 3 );
	ListNode* pNode5 = Find( pHead, 5 );

	pNode5->m_pNext = pNode3;

	cout << IsRing( pHead );

	return 0;



void push_back( ListNode** pHead, int value )

	if ( NULL == pHead )
		return;

	ListNode* pNewNode = new ListNode( NULL, value );

	if ( NULL == *pHead )
		*pHead = pNewNode;
		return;
	

	ListNode* pNode = *pHead;

	while ( NULL != pNode->m_pNext )
		pNode = pNode->m_pNext;

	pNode->m_pNext = pNewNode;


ListNode* Find( ListNode* pListHead, int value )

	if ( NULL == pListHead )
		return NULL;

	ListNode* pNode = pListHead;

	while ( NULL != pNode && value != pNode->m_nKey )
		pNode = pNode->m_pNext;

	if ( NULL == pNode )
		return NULL;

	return pNode;


//以后写代码，一定不要忘了考虑边界情况
bool IsRing( ListNode* pListHead )

	if ( NULL == pListHead )
		return false;

	ListNode* pFast= pListHead;
	ListNode* pSlow= pListHead;

	while ( NULL != pFast->m_pNext )
		if ( NULL != pFast->m_pNext->m_pNext )
			pFast = pFast->m_pNext->m_pNext;
			pSlow = pSlow->m_pNext;
		
		else
			return false;

		if ( pSlow == pFast )
			return true;
	

	return NULL;