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Question

c++怎么历遍输出带头结点的循环单链表的所有结点，用递归算法怎么解决，麻烦给给思路，谢谢

Answer 1

记录头结点位置，当节点循环到头结点的时候结束递归即可。

代码如下：

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

//递归显示
void showList(LinkNode* head)
{
    static LinkNode* node = 0;
    if(node == 0)
    {
        node = head;
        head = head->next;
    }
    if(node == head)
        return;
    else
    {
        if(head == node->next)
            cout << head->data;
        else
            cout << " "<<head->data;
        head = head->next;
        showList(head);
    }
}



int main()
{
    LinkNode* head,*p,*t;
    int i,n;
    head = new LinkNode;
    head->next = 0;

    p = head;
    //输入n
    cin >> n;
    for(i=0;i<n;i++)
    {
        t = new LinkNode;
        t->next = 0;
        cin >> t->data; //输入数据
        p->next = t;
        p = t;
    }
    //构建循环链表
    p->next = head; //尾结点链接到头结点
    showList(head);
    return 0;
}