單向連結串列環問題的全面解析（判斷環，環的長度，環的入口）

參照部落格https://my.oschina.net/u/2360415/blog/741253

#include <iostream>
using namespace std;

struct Node
{
	Node(int x)
	{
		value = x;
		next = 0;
	}
	int value;
	Node *next;
};

int getListLength(Node* head)
{
	Node *slow = head;
	Node *fast = head;

	while (fast&&fast->next)
	{
		slow = slow->next;
		fast = fast->next->next;
		if (fast == slow)
			break;
	}//如果有環，迴圈結束後，slow=fast是碰撞點--------------------------------

	//因為沒有環的情況下，fast有可能停留在NULL位置，也有可能停留在最後一個節點處
	//得加上fast->next==NULL
	if (fast == NULL||fast->next==NULL)//無環
	{
		int len = 0;
		Node *p = head;
		while (p)
		{
			len++;
			p=p->next;
		}
		//省略
		return len;
	}

	else
	{
		int len1 = 0, len2 = 0;//len1是頭節點到入口點的節點數(不含入口節點),len2是環所有的節點數
		Node *p1 = slow;
		Node *p2 = head;
		while (p1)
		{
			len1++;
			p1 = p1->next;
			p2 = p2->next;
			if (p1 == p2)
				break;

		}//迴圈結束後p1=p2是環的入口節點-------------------------------

		while (slow)//固定碰撞點fast指標
		{
			len2++;
			slow = slow->next;
			if (slow == fast)
				break;
		}
		cout << "環長：" << len2 << endl;
		return len1 + len2;
	}
}


int main()
{
	Node *p1 = new Node(1);
	Node *p2 = new Node(2);
	Node *p3 = new Node(3);
	Node *p4 = new Node(4);
	Node *p5 = new Node(5);
	Node *p6 = new Node(6);

	p1->next = p2;
	p2->next = p3;
	p3->next = p4;
	p4->next = p5;
	p5->next = p6;
	p6->next = p3;//節點p3是環入口點
	cout << getListLength(p1) << endl;

	return 0;
}