LRU算法简单实现

Posted 2020-07-12 鸣灭

package two;

import java.util.Scanner;

public class LRU {

    
    
    public static void main(String[] args) {
        int l=0;
        Integer []now=new Integer[5];
//        System.out.print(now[2]);
        System.out.println("输入多少个数值：");
        Scanner in=new Scanner(System.in);
        int count=in.nextInt();
        int []a=new int[count];
        System.out.println("输入他们");
        for(int i=0;i<count;i++){
            a[i]=in.nextInt();
        }
        
        boolean linshi=true;
        for(int i=0;i<count;i++){
            
            
            
                
            for(int j=0;j<l;j++){
                if(now[j]==a[i]){
                    for(int o=j;o>0;o--){
                        now[o]=now[o-1];
                        
                        
                        }
                        
                    now[0]=a[i];
                    linshi=false;
                   
                    break;
                }
            }
        
            if(linshi){
                for(int j=0;j<5;j++)
                    if(now[j]==null){
                        
                        for(int o=j;o>0;o--)
                            now[o]=now[o-1];
                        now[0]=a[i];
                        linshi=false;
                        l++;
                        break;
                    }
             }
            if(linshi){
                System.out.println("删除的值是"+now[4]);
                for(int j=4;j>0;j--){
                    
                    now[j]=now[j-1];
                }
                now[0]=a[i];
            }
            
            for(int j=0;j<now.length;j++){
                if(now[j]!=null){
                    
                    System.out.print(now[j]+"  ");
                    
                }
            }
            System.out.println();
            
            linshi=true;
         }
        
        
        }
//
}