200109题（vivo）

约瑟夫环问题的循环链表解法

最新推荐文章于 2023-06-07 22:45:37 发布

原创最新推荐文章于 2023-06-07 22:45:37 发布 · 204 阅读

0 ·

CC 4.0 BY-SA版权

本文详细介绍了使用循环链表解决约瑟夫环问题的两种方法，通过具体的C++代码实现，展示了如何创建链表、遍历并删除节点，以找出最后的幸存者。

在这里插入图片描述

思路：看到约瑟夫环我就喜欢用循环链表。

#include<iostream>
using namespace std;
typedef struct _node
{
	int num;
	struct _node * next;
	_node(int x) :num(x), next(NULL) {}
}node;

node*create(int N) {
	node*head =new node(1);
	node*prev = head;
	for (int i = 2; i <= N; i++)
	{
		node*curr = new node(i);
		prev->next = curr;
		prev = curr;
	}
	prev->next = head;
	return head;
}

void solution(int N, int M)
{
	if (N == 0)
		return;
	int count = 1;
	node*head = create(N);
	node*curr = head;
	if (M == 1)
	{
		while (N--)
		{
			cout << curr->num << " ";
			curr = curr->next;
		}
		return;
	}
	while (curr->next != curr)
	{
		count++;
		node*pnext = curr->next;
		if (count % M == 0)
		{
			cout << pnext->num << " ";
			curr->next = pnext->next;
			delete pnext;
		}
		else
		{
			curr = curr->next;
		}
	}
	cout << curr->num << " ";
	

}
int main() {

	//solution(6, 1);
	solution(6,3);
	system("pause");
	return 0;
}

牛客上有人的另一种解法，我觉得也不错：

void solution(int N, int M)
{
	node* head = new node(1);
	node* curr = head;
	for (int i = 2; i <= N; ++i) {
		curr->next = new node(i);
		curr = curr->next;
	}
	curr->next = head;
	node* prev = curr;
	curr = head;
	for (int i = 1; i <= N; ++i) {//N轮，每轮删一个结点
		for (int j = 1; j < M; ++j) {
			prev = curr;
			curr = curr->next;
		}
		cout << (curr->num) << " ";
        delete curr;
		prev->next = curr->next;
		curr = curr->next;
	}
}