约瑟夫环问题一解

问题描述：
约瑟夫环问题(Josephus)

用户输入M,N值，从1至N开始顺序循环数数，每数到M输出该数值，直至全部输出。写出C程序。（约瑟夫环问题 Josephus）

解法思路：

建立一个有N个元素的循环链表，然后从链表头开始遍历并记数，如果计数i==m(i初始为1)踢出元素，继续循环，当当前元素与下一元素相同时退出循环。

代码：

#include <iostream>
using namespace std;

typedef int Item;
typedef struct node * plist;

typedef struct node {
	Item key;
	node * next;
} node;
plist CreateList(int N){
	/* plist head,curr,tail;*/
	plist head,curr,tail;
	int i;
	head = NULL;
	for(i=1;i<=N;i++){
		curr = (plist)malloc(sizeof(node));
		curr->key = i;
		curr->next = NULL;
		if(!head){
			head = curr;
			tail = curr;
		} else{
			tail->next = curr;
			tail = tail->next;

		}
	}
	tail->next = head;
	return head;
};

void DisplayList(plist lst){
	plist curr = lst;
	do{
		printf("%d\t",curr->key);
		curr = curr->next;
	}while(curr!=lst);
}

void DestroyList(plist lst){
	plist curr = lst;
	if(!lst) 
		return;
	plist nxt;
	do{
		nxt = curr->next;
		free(curr);
		curr = nxt;
	}while(curr!=lst);
}
void Josephus(plist lst,int M){
	if(!lst)
		return;
	plist curr,pre;
	int i = 1;
	curr = lst;
	pre = lst;
	while(curr){
		if(i==M){
			printf("%d\t",curr->key);
			pre->next = curr->next;
			free(curr);
			curr = pre->next;
			i = 1;
		}
		pre = curr;
		curr = pre->next;
		if(pre == curr){
			printf("%d\n",curr->key);
			free(curr);
			break;
		}
		i++;

	}

}
void main()
{
	plist lst = CreateList(7);
	DisplayList(lst);
	printf("\nJosephus:\n");
	//DestroyList(lst);
	Josephus(lst,2);
}

输出结果：

1	2	3	4	5	6	7	
Josephus:
2	4	6	1	5	3	7