判断链表是否带环,若带环,找到环的入口点
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#pragma once #includeusing namespace std; template struct LinkNode { T _data; LinkNode* _next; LinkNode(const T& x) :_data(x) , _next(NULL) {} }; template class ListNode { //为了安全性 private: ListNode(const ListNode& l) {} ListNode & operator=(ListNode l) {} public: //程序限制 LinkNode * _head; public: ListNode() :_head(NULL) {} ~ListNode() { while (_head) { PopBack(); } } void PushBack(const T& x) { LinkNode * tmp = new LinkNode (x); if (_head == NULL) _head = tmp; else { LinkNode * cur = _head; while (cur->_next) cur = cur->_next; cur->_next = tmp; } } void PopBack() { if (_head == NULL) return; if (_head->_next == NULL) { delete _head; _head = NULL; } else { LinkNode * cur = _head; while (cur->_next&&cur->_next->_next) { cur = cur->_next; } LinkNode * del = cur->_next; cur->_next = NULL; delete del; } } LinkNode * Search(const T& x) { if (_head == NULL) return NULL; LinkNode * cur = _head; while (cur->_data != x) cur = cur->_next; return cur; } }; //判断链表是否带环 template bool CheckIsCircle(LinkNode * head) { if (head == NULL || head->_next == NULL) return false; LinkNode * fast, *slow; fast = slow = head; while (fast&&fast->_next) { fast = fast->_next->_next; slow = slow->_next; if (fast == slow) return true; } return false; } template LinkNode * SearchCircleAccess(LinkNode * head) { if (head == NULL||head->_next==NULL) return NULL; LinkNode * fast = head; LinkNode * slow = head; while (fast&&fast->_next) { fast = fast->_next->_next; slow = slow->_next; if (fast == slow) break; } slow = head; //于是我们从链表头、与相遇点分别设一个指针, //每次各走一步,两个指针必定相遇,且相遇第一点为环入口点。 //一个从头走,另一个从相遇点开始在环里走, //快指针比慢指针少走x,当它们相遇的第一个节点就是入口点 while (slow != fast) { slow=slow->_next; fast = fast->_next; } return slow; } void Test1() { ListNode l1; l1.PushBack(1); l1.PushBack(2); l1.PushBack(3); l1.PushBack(4); l1.PushBack(5); l1.PushBack(6); l1.PushBack(7); l1.PushBack(8); l1.PushBack(9); (l1.Search(9))->_next = l1.Search(6);//构建环 if (CheckIsCircle(l1._head)) { cout << (SearchCircleAccess(l1._head))->_data << endl; } }
运行结果:找到的入口点的数据为6
本文题目:判断链表是否带环,若带环,找到环的入口点
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