题解 | #链表中环的入口结点#

• 快慢指针相遇，判断有环
• 同时移动头节点和相遇时的快指针，再次相遇时，头节点便移动到了环入口处

        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
            if (slow == fast) {
                ListNode* index1 = fast;
                ListNode* index2 = pHead;
                while (index1 != index2) {
                    index1 = index1->next;
                    index2 = index2->next;
                }
                return index2;
            }
        }