剑指offer-面试题5 反向输出链表值

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typedef struct ListNode* List;struct ListNode{    int Value;    struct ListNode* Next;};void ReverseList1(List list)   //使用递归{    if(list != NULL)    {        if(list->Next != NULL)        {            ReverseList(list->Next);        }        std::cout<<list->Value<<" ";    }}void ReverseList2(List list)   //使用栈{    std::stack<List> stackList;    struct ListNode* node = list;    while(node != NULL)    {        stackList.push(node);        node = node->Next;    }    while(!stackList.isEmpty())    {        std::cout<<stackList.top()->Value<<" ";        stackList.pop();    }}void ReverseList3(List list)    //反转指针指向{    List p = NULL;    List temp = NULL;    if(list != NULL)    {        return NULL;    }else    {        temp = list->Next;        while(temp->Next != NULL)        {            p = temp->Next;            temp->Next = p->Next;            p->Next = list->Next;            list->Next = p;        }        while(list != NULL)        {            std::cout<<list->Value<<" ";            list = list->Next;        }    }}//反转链表ListNode* ReverseList2(ListNode* pHead){    ListNode* pNode=pHead;//当前结点    ListNode* pPrev=NULL;//当前结点的前一个结点    while(pNode!=NULL)    {        ListNode* pNext=pNode->m_pNext;        pNode->m_pNext=pPrev;//当前结点指向前一个结点        pPrev=pNode;//pPrev和pNode往前移动。        pNode=pNext;//这里要使用前面保存下来的pNext，不能使用pNode->m_pNext    }    return pPrev;//返回反转链表头指针。}
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