鏈表的逆序實現

題目：輸入一個鏈表的頭結點，反轉該鏈表。鏈表結點定義如下：

struct ListNode
{
      void*       m_nKey;
      ListNode* m_pNext;
};

常規實現，需要兩個臨時節點：

 ListNode* ReverseIteratively(ListNode* pHead)
{
      ListNode* pReversedHead = NULL;
      ListNode* pNode = pHead;
      ListNode* pPrev = NULL;
      while(pNode != NULL)
      {
            // get the next node, and save it at pNext
            ListNode* pNext = pNode->m_pNext;

            // if the next node is null, the currect is the end of original
            // list, and it's the head of the reversed list
            if(pNext == NULL)
                  pReversedHead = pNode;

            // reverse the linkage between nodes
            pNode->m_pNext = pPrev;

            // move forward on the the list
            pPrev = pNode;
            pNode = pNext;
   }

 

遞迴實現(不需要臨時節點)：

ListNode* reverse_list( ListNode* head)       //逆序
{
  ListNode* new_head=head;
 if(head==NULL || head->next==NULL)
  return head;
 new_head = reverse_list(head->next);
 head->next->next=head;
 head->next=NULL; //防止鏈表成為一個環，這是最關鍵的。
 return new_head; 
}

 

以上代碼都經過測試。