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单链表逆置

链表nizhi

题目:输入一个单向链表,将该单链表逆置。

举例:原来链表为1->2->3->4->5翻转为5->4->3->2->1

链表结点定义如下:

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

答:

#include "stdafx.h"
#include <iostream>
#include <fstream>

using namespace std;

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

//构造链表
void CreateList(ListNode *&pHead)
{
    fstream fin("list.txt");
    ListNode *pNode = NULL;
    ListNode *pTmp = NULL;
    int data;
    fin>>data;
    while (data)
    {
        pNode = new ListNode;
        pNode->m_nKey = data;
        pNode->m_pNext = NULL;
        if (NULL == pHead)
        {
            pHead = pNode;
            pTmp = pNode;
        }
        else
        {
            pTmp->m_pNext = pNode;
            pTmp = pNode;
        }

        fin>>data;
    }
}

//翻转单链表
void ReverseLink(ListNode *&pHead)
{
    if (NULL == pHead)
    {
        return;
    }
    ListNode *pNode = pHead;
    ListNode *Prev = NULL;
    ListNode *pNext = NULL;
    while (NULL != pNode)
    {
        pNext = pNode->m_pNext;
        if (NULL == pNext)
        {
            pHead = pNode;
        }
        pNode->m_pNext = Prev;
        Prev = pNode;
        pNode = pNext;
    }
}

void PrintList(ListNode *pHead)
{
    if (NULL == pHead)
    {
        return;
    }
    ListNode *pNode = pHead;
    while (NULL != pNode)
    {
        cout<<pNode->m_nKey<<"  ";
        pNode = pNode->m_pNext;
    }
    cout<<endl;
}

int _tmain(int argc, _TCHAR* argv[])
{
    ListNode *pHead = NULL;
    cout<<"原来的链表:";
    CreateList(pHead);
    PrintList(pHead);
    ReverseLink(pHead);
    cout<<"翻转的链表:";
    PrintList(pHead);

    return 0;
}

运行界面如下:

建造链表的list.txt文件如下:

12 11 10 9 8 7 6 5 4 3 2 1 0
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