问题描述:
约瑟夫(Josephus)问题:设有n 个人围成一个圆圈,任意给定一个正整数m,从第一个人开始顺时针计数,计到第m个人,将其从圆圈中除去。然后再从下一个人开始,周而复始,直到圆圈中只剩一个人为止,那么剩下的那个人就是赢家。
代码实现:
#include <assert.h> #include <iostream> using namespace std; struct Node { int Data; struct Node* next; }; Node* CreateCircleList() { Node* head = new Node; assert(head); head->next = NULL; int data; bool bInPuts = true; Node* q = head; while (bInPuts) { cin>>data; if (0 == data) { bInPuts = false; } else { Node* p = new Node; assert(p); p->Data = data; p->next = head; q->next = p; q = p; } } return head; } int GetLength(Node* head) { assert(head); int num = 0; Node* p = head->next; while (p != head) { num++; p = p->next; } return num; } /** * @brief 该函数用来解决“约瑟夫问题” * @return 返回循环链表的头指针 * @param[in] head 约瑟夫链 * @param[in] m 计数值 * @return 返回获胜者元素值 * @author wlq_729@163.com * http://blog.csdn.net/rabbit729 * @version 1.0 * @date 2009-03-10 */ int Josephus(Node* head, const int m) { assert(head); if (head->next == head) { cout<<"link is null!"<<endl; return -1; } int count = 0; Node* p = head; Node* prior = head; while (GetLength(head) > 1) { // 寻找待删除结点,p为待删除结点,prior为其前驱结点 while (count < m) { count++; prior = p; p = p->next; // 考虑尾指针情况 if (p == head) { prior = head; p = head->next; } } // 删除结点 cout<<"输出:"<<p->Data<<endl; prior->next = p->next; count = 0; } return head->next->Data; } void Show(Node* head) { assert(head); if (head->next == head) { cout<<"link is null!"<<endl; return; } Node* p = head->next; while (p != head) { cout<<p->Data<<endl; p = p->next; } return; } void main(void) { // 建立循环链表 Node* head = CreateCircleList(); // 显示链表 Show(head); // 求取获胜者 int e = Josephus(head, 3); cout<<"获胜者为:"<<e<<endl; }
程序输出结果:
假设链表为1,2,3,4,5,6。
则程序运行输出为:
123456输出:3输出:6输出:4输出:2输出:5获胜者为:1请按任意键继续. . .
posted on 2009-03-10 21:29 张云临 阅读(195) 评论(0) 编辑 收藏 举报