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                ### 這道題是典型的用快慢指針解決的:慢指針每次走一步,快指針每次走兩步,如果它們能相遇(快指針趕上慢指針),則存在環,如果遇到了nullptr,則不存在環。 ``` // 如果存在環,則返回相遇的點,否則返回nullptr SList* meetingNode(SList *list) { // 邊界條件:鏈表為空 if (list == nullptr) { return nullptr; } SList *slow = list; if (slow == nullptr) { return nullptr; } SList *fast = slow->m_next; while (fast != nullptr && slow != nullptr) { if (fast == slow) { return fast; } slow = slow->m_next; fast = fast->m_next; if (fast != nullptr) { fast = fast->m_next; } } // 遇到nullptr return false; } ```
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