题解 | #链表中环的入口节点#

用快慢指针，分别从两指针相遇位置和头指针位置出发两个相同速度的指针，相遇的位置就是环开始的位置。

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode *fast = head, *slow = head, *pos = nullptr;
        while(fast != nullptr && fast->next != nullptr)
        {
            fast = fast->next->next;
            slow = slow->next;
            if(fast == slow) 
            {
                pos = fast;
                break;
            }
        }
        if(pos == nullptr) return pos;
        else
            while(head != pos)
            {
                head = head->next;
                pos = pos->next;
            }
        return pos;
    }
};