Recursive solution number of nodes in the single list

#include <iostream>
#include<iomanip>
using namespace std;
#define MAXSIZE 10000

typedef struct LNode
{
	double data;
	struct LNode *next;
}LNode,*LinkList;

void InitList(LinkList &L)
{
	L=new LNode;
	L->next=NULL;
	return;
}

void CreateList_R(LinkList &L,int n)
{
	LinkList r,p;int i=0;
	L=new LNode;
	L->next=NULL;
	r=L;
	
	//int n;
	//cin>>n;
	
	while(1) 
	{
		p=new LNode;
		cin>>p->data;
		
		p->next=NULL;
		r->next=p;
		r=p;
		i++;
		if(i==n) break;
	}
	return;
}

int ListLength(LinkList L)
{
    LinkList p;
    p=L->next;  //p指向第一个结点
    int i=0;             
    while(p){//遍历单链表,统计结点数
           i++;
		   cout<<p->data<<endl;  
           p=p->next;	    
    } 
    return i;                             
 }
 
double Sum(LinkList p,double sum)
{
 	if(p->next!=NULL)
 	{
 		sum = sum + 1;
        return Sum(p->next, sum);
	}
 	else
	{
        return sum+1;
	}
}

int main()
{
	LinkList L;int n;int i=0;double sum=0;//未赋值导致不A 
	InitList(L);
	while(1)
	{
		cin>>n;
		if(n==0)break;
		CreateList_R(L,n);	
		cout<<Sum(L->next,sum)<<endl;
	}
	return 0;
}

description

 

Using a single linked list represents a sequence of integers, the recursive method of calculating the number of nodes in a single linked list.

Entry

A plurality of sets of data, each data has two rows, the first row of chain length n, the second line of the list of n elements (separated by a space between the elements). When n = 0, the input end.

Export

For each data output line, respectively, corresponding to the number of each node in the linked list.

Sample input 1 

4
1 2 3 4
6
1 2 43 5 7 2
0

Sample Output 1

4
6