1007. Find the middle element in a Linked List

Description

Assuming the following definition:

template <typename Node_entry>struct Node {
   Node_entry entry;
   Node *next;
   Node();
   Node(Node_entry item, Node *add_on = NULL);
};

template <typename Node_entry> Node<Node_entry>::Node()
{
   next = NULL;
}

template <typename Node_entry> Node<Node_entry>::Node(Node_entry item, Node *add_on)
{
   entry = item;
   next = add_on;
}

Implement the following function:
template<typename T> const Node<T>* medianElement( const Node<T> *  head )

//Precondition: the linked list head is not empty.（但是需要判断为头空时的情况！）
//Postcondition: it returns the pointer to the middle element in the linked list head.
//If there are even number of elements in the list, the first one of the two middle elements is returned. 
 

For example,

 

在单链表中找到中间的元素。

#include <iostream>
using namespace std;

template <typename Node_entry>struct Node {
    Node_entry entry;
    Node *next;
    Node();
    Node(Node_entry item, Node *add_on = NULL);
};

template <typename Node_entry> Node<Node_entry>::Node()
{
    next = NULL;
}

template <typename Node_entry> Node<Node_entry>::Node(Node_entry item, Node *add_on)
{
    entry = item;
    next = add_on;
}

template<typename T> const Node<T>* medianElement( const Node<T> *  head ) {
    int count = 0;
    if (head == NULL) {
        return head;
    }
    const Node<T>* temp = new Node<T>();
    temp = head;
    
    //计算链表的长度
    while (temp != NULL) {
        count++;
        temp = temp->next;
    }
    
    //计算出中位数
    int half = 1;
    if (count % 2 == 0) {
        half = count / 2;
    }
    else half = count / 2 + 1;
    
    //获得元素
    while (half != 1) {
        head = head->next;
        half--;
    }
    return head;
}