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                ### 說明:無環鏈表。這道題也是歸并排序的中間步驟。 ### 思路:這道題思路和[倒數第 k 個節點](http://www.hmoore.net/persuez/algorithm/861500)的思路是一致的,都是用距離是固定的,只不過這道題用一個特別一點的方式求:快指針每次走兩步到最后的或倒數第一個節點(節點數為偶數,都是從鏈頭出發),慢指針每次走一步到達的就是中間節點(偶數的話是左節點)。 以下是代碼實現: ``` SList* findMiddleNode(SList *pHead) { if (pHead == nullptr) { return nullptr; } SList *slow = pHead, *fast = pHead; while (fast != nullptr && fast->m_next != nullptr) { fast = fast->m_next->m_next; slow = slow->m_next; } return slow; } ```
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