题解 | #链表中环的入口结点#

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
        val(x), next(NULL) {
    }
};
*/
class Solution {
public:
    ListNode* EntryNodeOfLoop(ListNode* pHead) {
        //判断环并找出入口
        ListNode* fast = pHead;
        ListNode* slow = pHead;
        while(fast != nullptr && fast->next != nullptr)
        {
            //一个走一步一个走两步，如果有环一定相遇
            fast = fast->next->next;
            slow = slow->next;
            //相遇了跳出
            if(fast == slow) break;
        }
	  //判断是不是break跳出的循环，是则有环
        if(fast == nullptr || fast->next == nullptr) return nullptr;
	  //head到入口的距离等于相遇点到入口的距离加上(n)个环的长度r
        slow = pHead;
        while(slow != fast)
        {
            fast = fast->next;
            slow = slow->next;
        }
        return fast;
    }
};