题目：

#include <iostream>
#include <vector>
using namespace std;
/*
 Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.
*/

typedef struct list_node List;
struct list_node
{
	struct list_node* next;
	int value;
};

void print_list(List* list)
{
	List* tmp=list;
	while(tmp != NULL)
	{
		cout<<tmp->value<<endl;
		tmp = tmp->next; 
	}
}

/*
初始化List  将从1~n的数字插入到链表中 
*/
void Init_List(List*& head,int* array,int n)
{
	head = NULL;
	List* tmp;
	List* record;
	 
	for(int i=1;i<=n;i++)
	{
		tmp = new List;
		tmp->next = NULL;
		tmp->value = array[i-1];
		if(head == NULL)
		{
			head = tmp;
			record = head;
		}
		else
		{
			record->next = tmp;
			record = tmp;
		}
	}
}

int Len_list(List* list)
{
	if(list == NULL)
		return 0;
	else
		return Len_list(list->next)+1;
}

/*
重新排序链表  将一个链表拆分 然后再重新组合 
具体的情况也得分析来看  如果节点数目是偶数个  那么正好平分
然后将后一半的链表反转 之后就可以插入了
如果节点数据是奇数个  那么前半部多一个  然后后半部反转
之后进行插入 
*/
/*
链表的反转 
*/
void Reverse(List*& list)
{
	List* tmp = NULL;
	List* cur = list;
	List* next = list->next;
	while(next != NULL)
	{
		cur->next = tmp;
		tmp = cur;
		cur = next;
		next = next->next;
	} 
	cur->next = tmp;
	list = cur;
}

void Reorder_list(List*& list)
{
	List* first = list;
	List* second;
	List* tmp_first,*tmp_second;
	int len = Len_list(first);
	int i;
	if(len%2 == 0)
	{
		for(i=1;i<len/2;i++)
			first = first->next;
	}
	else
	{
		for(i=1;i<(len/2)+1;i++)
			first = first->next;
	} 
	second = first->next;
	first->next = NULL;
	Reverse(second);
	
	first = list;
 // 开始进行融合 首先可以保证的是second链表的个数肯定不会比first链表的节点数目多
	while(second != NULL)
	{
		tmp_first = first->next;
		tmp_second = second->next;
		first->next = second;
		second->next = tmp_first;
		second = tmp_second;
		first= tmp_first;
	} 
	
}

int main() 
{
	int array[]={1,2,3,4,5,6,7,8,9,10,11,12};
	List* head;
	Init_List(head,array,sizeof(array)/sizeof(array[0]));
 
	Reorder_list(head);
	print_list(head);
	return 0;
}