[小算法] 找出单链表中的中间元素

两种情况:

1. 链表有奇数个结点， 中间元素只有一个；

2. 链表有偶数个结点， 中间元素会有两个；

 

#include "stdafx.h"
#include <iostream>
using namespace std;

struct Link
{
    int Data;
    Link *Next;
};

void PrintMiddleNode(Link *head);

int _tmain(int argc, _TCHAR* argv[])
{
    struct Link head, node1,node2,node3,node4,node5;
    head.Data = -1;
    head.Next = &node1;
    node1.Data = 1;
    node1.Next = &node2;
    node2.Data = 2;
    node2.Next = &node3;
    node3.Data = 3;
    node3.Next = &node4;
    node4.Data = 4;
    node4.Next = &node5;
    node5.Data = 5;
    node5.Next = NULL;

    PrintMiddleNode(&head);
    cout << endl;

    cin.get();
    return 0;
}

void PrintMiddleNode(Link *head)
{
    if(head == NULL || head->Next == NULL) return;
    Link *singleStepPointer = head;
    Link*doubleStepPointer = head;

    while(true)
    {
        if(doubleStepPointer->Next != NULL)
        {
            doubleStepPointer = doubleStepPointer->Next;
            if(doubleStepPointer->Next != NULL)
            {
                doubleStepPointer = doubleStepPointer->Next;
            }
            else//奇数个
            {
                //middle node should be singleStepPointer->Next
                cout << "Middle node is: " << singleStepPointer->Next->Data;
                break;
            }
        }
        else//偶数个
        {
            //middle nodes should be current singleStepPointer and singleStepPointer->Next
            cout << "Middle nodes are: " << singleStepPointer->Data << " and " << singleStepPointer->Next->Data;
            break;
        }

        singleStepPointer = singleStepPointer->Next;
    }
}