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                ### 說明:無環鏈表。這道題個人感覺只要思路清晰就可以做出來,重點是定義 3 個指針,保存的內容為:前一個節點,當前要將m_next指向前一個節點的節點,后一個節點。不斷向前遍歷。注意點是第一個節點的 m_next 要指向空指針,頭指針要指向最后的那個節點。 ``` SList* reverseList(SList *pHead) { SList *pReverseHead = nullptr; // 保存反轉后的頭節點 // 定義 3 個指針 SList *pPrev = nullptr, *pCur = pHead, *pNext = nullptr; while (pCur != nullptr) { // 保存下個節點到pNext中 pNext = pCur->m_next; // 保存反轉后的頭節點 if (pNext == nullptr) { pReverseHead = pCur; } // 設置當前節點的m_next pCur->m_next = pPrev; // 向前走 pPrev = pCur; pCur = pNext; } // 返回鏈表頭 return pReverseHead; } ```
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