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) {

        ListNode* fast = pHead;
        ListNode* slow = pHead;

        while(fast && fast->next){
            fast = fast->next->next;
            slow = slow->next;
            if(fast == slow){
                // 通过分析知道，从head开始到入口的距离等于从相遇点到入口的距离
                ListNode* meet = fast;
                ListNode* head = pHead;
                while(head != meet)
                {
                    meet = meet->next;
                    head = head->next;
                }
                return head;
            }
        }
        return nullptr;
    }
};