就如数字6一样的单链表结构，如何检测是否有6下部的○呢，并且求交叉点位置

思路

1. 使用快慢指针（一个一次走2步，一个走1步），若快慢指针第一次相遇，则有环
   慢指针路程 $s=a+b$
   快指针路程 $2s=a+b+n\ast R$
   $s=n\ast R$
   链表的长度 $L=a+b+c=n\ast R+c$
   又 $L=a+R$
   则 $n\ast R+c=a+R\Rightarrow (n-1)\ast R+c=a$
   意味着从head开始走到入口 $a$ 步 == 从第一次相遇点走 $n-1$ 圈 + 再走 $c$ 步

2. 由上可以使快慢指针第一次到达相遇点时，使慢指针回到head，快指针仍在相遇点，然后两人步伐一致，最后会在入口相见。

C++代码实现：

完整代码见：https://github.com/hitskyer/course/tree/master/dataAlgorithm/chenmingming/linkedList

类实现函数

bool SingleList::hasLoop()
{bool loop = false;ListNode fast = m_pHead, slow = m_pHead;ListNode posMeet = m_pHead, ringEntrance = m_pHead;//环的可能的入口应初始化成headif(m_pHead == NULL){loop = false;std::cout << "list has no loop!" << std::endl;}else{while(fast && fast->pNext){fast = fast->pNext->pNext;slow = slow->pNext;if(fast == slow){loop = true;posMeet = fast; //第一次相遇的地方break;}}slow = m_pHead; //接着让慢指针回到表头（这里是关键），继续一起同步前行，第二次相遇的地方为环的入口while(slow != fast){slow = slow->pNext;fast = fast->pNext;if(fast == slow)ringEntrance = fast;}size_t lenOf_headToEntrance = howManyNode(m_pHead,ringEntrance);size_t ringLen_1 = howManyNode(ringEntrance->pNext, ringEntrance);std::cout << "len of head to ring entrance is " << lenOf_headToEntrance << std::endl;std::cout << "entrance Node is " << ringEntrance->data << std::endl;std::cout << "len of ring is " << ringLen_1 + 1 << std::endl;std::cout << "len of List is " << lenOf_headToEntrance + ringLen_1 + 1 << std::endl;}return loop;
}size_t SingleList::howManyNode(ListNode ps, ListNode pe)	//计算两个指针之间有多少个节点（不包含第二个参数处的节点）
{size_t count = 0;while(ps != pe){ps = ps->pNext;++count;}return count;
}

singleListIsLoop.cpp主函数

//
// Created by mingm on 2019/3/24.
//检查单链表中是否存在环，求环的长度，链表长度，及环的入口
#include <iostream>
#include <time.h>
#include <cstdlib>
#include "./homework/singleList.cpp"
using namespace std;
int main()
{srand((unsigned)time(NULL));    //用时间随机数种子size_t len = 10;       //测试链表最大长度for(size_t j = 1; j < len; ++j){SingleList intList;for(size_t i = 0; i < j; ++i){intList.AddTail(rand()%100);    //添加随机数到链表}cout << "no loop list: " << endl;intList.PrintList();    //排序前链表打印size_t n = rand()%(intList.GetLength());    //0-链表减1的随机数ListNode randNode = intList.GetHeadNode();for(size_t i = 0; i != n; ++i){randNode = randNode->pNext;     //链表的一个随机节点}ListNode originTail = intList.GetTailNode();originTail->pNext = randNode;    //尾节点接入链表中的随机位置形成环intList.hasLoop();  //调用环检测函数originTail->pNext = NULL;   //断开环，让链表能够按照单链表析构函数析构!!!!!!!std::cout << "getListLength（） is " << intList.GetLength() << std::endl;std::cout << "-----------------------------------------" << std::endl;}return 0;
}

Valgrind检测结果

从链表长度1开始测试到9，测试结果均正确！