約瑟夫環
問題描述：
m個人圍成一個圈，指定一個數字n,從第一個人開始報數，每輪報到n的選手出局，由下一個人接着從頭開始報，最后一個人是贏家。其中m>1,n>2。
鏈表法
用循環鏈表能完美契合本題
#include<iostream>
using namespace std;
struct Node{
int data;
Node* next;
Node(int value){this->data = value;};
};
//創建一個約瑟夫環的類
class JosephCircle{
private:
Node* tail; //尾結點
//Node* eliminate; //淘汰結點
public:
JosephCircle():tail(NULL){}
~JosephCircle(){delete tail;}
void Add(int num);
void Eliminate(int step);
void Print();
};
void JosephCircle::Add(int num){
if(tail == NULL){
tail = new Node(num);
tail->next = tail;
}
else{
Node* new_node = new Node(num);
new_node->next = tail->next;
tail->next = new_node;
tail = new_node;
}
}
void JosephCircle::Eliminate(int step){
Node* p = tail;
while(p != NULL && p != p->next){
for(int i = 0;i < step-1;i++){
p = p->next;
}
Node* eliminate = p->next;
p->next = eliminate->next;
if(eliminate == tail){
tail = p;
}
cout<<"deleting"<<eliminate->data<<endl;
delete eliminate;
Print();
}
}
void JosephCircle::Print(){ //這打印還是有點說法的
Node* cur = tail;
while(cur != NULL){
cur = cur->next;
cout<<cur->data<<" ";
if(cur == tail)
break;
}
cout<<endl;
}
int main(){
JosephCircle jc;
for(int i = 1;i <= 6;i++){
jc.Add(i);
}
jc.Eliminate(3);
jc.Print();
return 0;
}
數組
數組倒是也能完成，代碼量好像還要少一丟丟，但是要注意的邊界條件太多了，debug的時間都夠我寫個鏈表解決的了T_T
#include <iostream>
using namespace std;
int JCArr(int num,int step){
int arr[num];
for (int i = 1; i <= num; ++i){
arr[i-1] = i;
}
int n = num,i = 0,curOut = 1;
while(num != 1){
if(arr[i] == -1){
i++;
}
else{
i++;
curOut++;
}
if(i == n){ //超過末尾從頭開始
i = 0;
}
if(curOut == step && arr[i] != -1){
cout<<"deleting "<<arr[i]<<endl;
arr[i] = -1;
i++;
if(i == n){ //這里也有可能發生越界，小心呀！
i = 0;
}
curOut = 1;
num--; //別忘了總人數減一
}
}
int k = 0;
for(;k < n;k++){
if(arr[k] != -1)
return arr[k];
}
}
int main(int argc, char const *argv[])
{
cout<<JCArr(6,3)<<endl;
return 0;
}
遞歸
//TO DO
參考鏈接：