Josephus问题的实现
作者:互联网
这里使用队列来解决。
因为我们处理的是n个元素里面的第m个元素,如果每次从队列里一边取元素,一边又加到队列的末尾,数到第m的时候,这第m的元素直接出队,不再入队。依此循环n遍,可以按所需顺序移除掉n个元素。
C++代码如下:
1 #include <iostream> 2 #include <stdlib.h> 3 using namespace std; 4 5 struct Node 6 { 7 int data; 8 struct Node* next; 9 }; 10 typedef struct Node* queue; 11 typedef struct Node* position; 12 //创建队列 13 queue create_queue(int N) 14 { 15 queue Q = (queue)malloc(sizeof(Node)); 16 Q->next = NULL; 17 position r=NULL; 18 for (int i = 1; i <= N; i++) 19 { 20 position p = (position)malloc(sizeof(Node)); 21 p->data = i; 22 if (Q->next == NULL) 23 Q->next = p; 24 else 25 r->next = p; 26 r = p; 27 } 28 r->next = NULL; 29 return Q; 30 } 31 //出队 32 position Dequeue(queue Q) 33 { 34 if (Q->next == NULL) 35 { 36 cout << "queue is empty" << endl; 37 exit(1); 38 } 39 else 40 { 41 position front = Q->next; 42 Q->next = front->next; 43 front->next = NULL; 44 return front; 45 } 46 } 47 //入队 48 void Enqueue(queue Q,position p) 49 { 50 position r=Q; 51 position rear = (position)malloc(sizeof(Node)); 52 if (rear == NULL) 53 { 54 cout << "out of space"; 55 exit(1); 56 } 57 while(r ->next!= NULL) 58 r = r->next; 59 r->next = rear; 60 rear->data = p->data; 61 rear->next = NULL; 62 free(p); 63 } 64 //Josephus 65 void Josephus(queue Q, int N,int M) 66 { 67 position p,r=Q; 68 for (int i = 0; i <N; i++)//循环N次 69 { 70 for (int j = 0; j < M-1; j++)//出队和入队M-1次 71 { 72 p=Dequeue(r);//出队 73 Enqueue(r,p);//入队 74 } 75 p = Dequeue(r);//出队第M次结点,但不再入队 76 cout<<p->data<<" "; 77 } 78 } 79 int main() 80 { 81 queue Q=create_queue(10); 82 Josephus(Q,10,5); 83 system("pause"); 84 return 0; 85 }
运行结果:
标签:Node,queue,Josephus,实现,next,问题,int,position,NULL 来源: https://www.cnblogs.com/cs0915/p/12152078.html