c++的相关问题，用c++的方法解决

关于输出有序链表的问题，从小到大依次建立1个带头结点，用c++的方法解决，会的给给思路

建表的时候，根据数据的大小查找插入的位置，调整前后两个节点的next指针即可。

如下：

#include <iostream>
using namespace std;
typedef struct _node
{
    int data;
    struct _node* next;
}LinkNode;

//插入链表
LinkNode* insertlist(LinkNode* head,LinkNode* t)
{
    LinkNode* p,*front;
    front = head;
    p = front->next;

    while(p && p->data < t->data)
    {
        front = p;
        p = p->next;
    }
    if(p)
    {
        front->next = t;
        t->next = p;
    }else
    {
        front->next = t;//插入最后
    }
    return head;
}


//创建链表
LinkNode* createList()
{
    int data;
    LinkNode *head,*p,*t;
    head = new LinkNode;
    head->next = 0;
    p = head;
    while(1)
    {
        cin >> data;
        if(data ==0)
            break;
        
        t = new LinkNode;
        t->data = data;
        t->next = 0;
    
        //t插入链表
        head = insertlist(head,t);


    }
    return head;
}

//分离链表
LinkNode* split(LinkNode* head)
{
    LinkNode* p,*front,*t;
    LinkNode* head2 = new LinkNode;
    head2->next = 0;
    t = head2;

    front = head;
    p = front->next;
    while(p)
    {
        if(p->data%2==0)
        {
            front->next = p->next;
            t->next = p;
            p->next = 0;
            t = p;

            p = front->next;
        }else
        {
            front = p;
            p = p->next;
        }
        
    }
    return head2;
}

//显示链表
void showlist(LinkNode* head)
{
    LinkNode* p = head->next;
    while(p)
    {
        if(p == head->next)
            cout << p->data;
        else
            cout <<"->"<<p->data;
        p = p->next;
    }
    cout <<endl;
}
//分隔链表


int main()
{
    LinkNode* head,*head1,*head2;
    head = createList();
    showlist(head);
    head2 = split(head);
    //显示
    showlist(head);
    showlist(head2);
    return 0;
}