使用顺序表求解约瑟夫环问题 （自定义顺序表）

约瑟夫环（Josephus）问题：古代某法官要判决n个犯人的死刑，他有一条荒唐的法律，将犯人站成一个圆圈，从第s个人开始数起，每数到第d个犯人，就拉出来处决，然后再数d个，数到的人再处决……直到剩下的最后一个可赦免。当n=5，s=1，d=2，时：

第一步：定义一个顺序表SeqList<T>：

public class SeqList<T> implements LList<T>{private Object[] element;private int len;public SeqList(int size){this.element=new Object[size];this.len=0;}public SeqList(){this(64);}public SeqList(SeqList<T> list){//拷贝构造方法（深度拷贝）this.len=list.len;this.element=new Object[list.element.length];for(int i=0;i<list.element.length;i++){this.element[i]=list.element[i];}}public boolean isEmpty(){return this.len==0;}public int length(){return this.len;}public T get(int i){if(i>=0&&i<this.len)return (T)this.element[i];return null;}public void set(int i,T x){if(x==null)return;if(i>=0&&i<this.len)this.element[i]=x;elsethrow new IndexOutOfBoundsException(i+"");}public void insert(int i,T x){if(x==null)return;if(this.len==this.element.length){Object[] temp=this.element;this.element=new Object[this.element.length*2];for(int j=0;j<temp.length;j++){this.element[j]=temp[j];}}if(i<0){i=0;}if(i>this.len){i=this.len;}for(int j=this.len-1;j>=i;j--){this.element[j+1]=this.element[j];}this.element[i]=x;this.len++;}public void append(T x){insert(this.len,x);}public T remove(int i){if(this.len==0||i<0||i>=this.len)return null;T old=(T)this.element[i];for(int j=i;j<this.len-1;j++){this.element[j]=this.element[j+1];}this.element[this.len-1]=null;this.len--;return old;}public void removeAll(){this.len=0;}public T search(T key){int index=this.indexOf(key);return index==-1?null:(T)this.element[index];}public int indexOf(T key){if(key!=null){for(int i=0;i<this.len;i++){if(this.element[i].equals(key)){return i;}}}return -1;}public boolean contain(T key){return this.indexOf(key)>=0;}public boolean equals(Object o){if(this==o)return true;if(o instanceof SeqList){SeqList<T> list=(SeqList<T>)o;if(list.length()==this.length()){for(int i=0;i<list.length();i++){if(!this.get(i).equals(list.get(i)))return false;}return true;}}return false;}public String toString(){String str="(";if(this.len>0){for(int i=0;i<this.len;i++){if(i<this.len-1)  str+=this.element[i].toString()+",";else  str+=this.element[i].toString();}}return str+")";}}

第二步：定义一个约瑟夫环并求解：

public class Josephus {public Josephus(int number,int start,int distance){SeqList<String> list=new SeqList<String>(number);for(int i=0;i<number;i++){list.append((char)('A'+i)+"");}System.out.print("约瑟夫环("+number+","+start+","+distance+"),");System.out.println(list.toString());int i=start;while(list.length()>1){i=(i+distance-1)%list.length();System.out.print("删除的元素："+list.remove(i).toString()+",");System.out.println(list.toString());}System.out.println("被赦免的罪犯是："+list.get(0).toString());}public static void main(String[] args) {new Josephus(5,0,2); }}

 

第三步运行结果：

约瑟夫环(5,0,2),(A,B,C,D,E)
删除的元素：B,(A,C,D,E)
删除的元素：D,(A,C,E)
删除的元素：A,(C,E)
删除的元素：E,(C)
被赦免的罪犯是：C