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!=NULL && fast->next != NULL) {
            fast = fast->next->next;
            slow = slow->next;

            if(slow == fast) {   // 找到相遇的节点
                break;
            }
        }

        if(fast == nullptr || fast->next == nullptr) {
            return nullptr;
        }

        fast = pHead;

        while(fast != slow) {    // 头结点和相遇点到环入口距离一致
            fast = fast->next;
            slow =slow->next;
        }

        return fast;
    }
};