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;
}
}