在Ｏ(1)时间删除链表结点

来源：互联网发布：C 先序遍历的递归算法编辑：程序博客网时间：2024/06/06 10:55

　给定单向链表的头指针和一个结点指针，定义一个函数在Ｏ(1)时间删除该结点．链表结点与函数定义如下：
　

struct ListNode{    int m_nValue;    ListNode* m_pNext;｝;

code:

//把待删结点后面一个结点的值赋给待删结点，然后把待删结点next指针指向下下个结点，然后删除下个结点，　达到和删除待删结点一样的效果．void DeleteNode(ListNode** pListHead, ListNode* pToBeDeleted){    if (!pListHead || !pToBeDeleted)        return;    //链表仅有一个结点就是待删结点    if (*pListHead == pToBeDeleted &&  pToBeDeleted->m_pNext == NULL)    {        delete pToBeDeleted;        pToBeDeleted = NULL;        *pListHead = NULL;    }    //链表不止一个结点并且要删除的结点是头节点    else if (*pListHead == pToBeDeleted)    {        *pListHead = (*pListHead)->m_pNext;        delete pToBeDeleted;        pToBeDeleted = NULL;    }   //要删除的结点是最后一个结点    else if (pToBeDeleted->m_pNext == NULL)    {        ListNode* pTmp = *pListHead;        while (pTmp->m_pNext != pToBeDeleted)        {            pTmp = pTmp->m_pNext;        }        pTmp->m_pNext = NULL;        delete pToBeDeleted;        pToBeDeleted = NULL;        pTmp = NULL;    }    //待删除结点为正常结点    else    {        ListNode* pNext = pToBeDeleted->m_pNext;        pToBeDeleted->m_nValue = pNext->m_nValue;        pToBeDeleted->m_pNext = pNext->m_pNext;        delete pNext;        pNext = NULL;    }}

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