本文实例为大家分享了java实现银行家算法的具体代码，供大家参考，具体内容如下
import java.util.Arrays;
import javax.swing.JOptionPane;
public class Banker_Dijkstra {
static int available[]={3,3,2}; //可利用资源数
static int max[][]={{7,5,3},{3,2,2},{9,0,2},{2,2,2},{4,3,3}};; //每线程大需求
static int allocation[][]={{0,1,0},{2,0,0},{3,0,2},{2,1,1},{0,0,2}}; //已分配资源
static int need[][]={{7,4,3},{1,2,2},{6,0,0},{0,1,1},{4,3,1}}; //需求
static int request[]=new int[3]; //存放请求资源
static int thread; //线程号
static JOptionPane jpane = new JOptionPane();
// static boolean m;
public static void main(String[] argv){
int n = 0 ;
Banker_Dijkstra bd = new Banker_Dijkstra();
for(int i=0;i<5;i++){
if(bd.safeState(i)){
JOptionPane.showMessageDialog(jpane, "系统状态安全");
n=1;
break;
}else{
n=2;
continue;
}
}
if(n==1){
bd.getThread();
}
else if(n==2){
JOptionPane.showMessageDialog(jpane, "系统状态不安全");
}
}
protected void getThread(){//输入测试线程号且输出结果
try{
String xiancheng = JOptionPane.showInputDialog(jpane,"请输入申请资源的线程:");
thread = (int) Integer.parseInt(xiancheng);
}catch(Exception e){
int response = JOptionPane.showConfirmDialog(jpane, "请输入0～4:",null, JOptionPane.YES_NO_OPTION);
// 处理异常
if(response==0){
getThread();
}else if(response ==1){
System.exit(0);
}
}
if(thread<0||thread>4){
JOptionPane.showMessageDialog(jpane, "请输入0～4:");
getThread();
}else{
for(int i=0;i<3;i++){
String requestR = JOptionPane.showInputDialog(jpane,"请输入申请的第"+(i+1)+"种资源(若不申请则填0)");
try{
request[i]=Integer.parseInt(requestR);}
catch(Exception e){
JOptionPane.showConfirmDialog(jpane, "请输入申请的第"+(i+1)+"种资源(若不申请则填0)",null,JOptionPane.YES_NO_OPTION);
}
}
if(request[0]>need[thread][0]||request[1]>need[thread][1]||request[2]>need[thread][2]){
JOptionPane.showMessageDialog(jpane,thread+"线程申请的资源超出其需要的资源，请重新输入");
getThread();
}else{
if(request[0]> available[0]||request[1]> available[1]||request[2]> available[2]){
JOptionPane.showMessageDialog(jpane,thread+"线程申请的资源大于系统资源，请重新输入");
getThread();
}
}
// 分配资源
allocateData(thread);
// 判断 继续模拟选择与处理
int tag=0;
if(check(thread)){
try{
String str = JOptionPane.showInputDialog(jpane,"是／否 继续模拟？( 1/0 ):");
tag = Integer.parseInt(str);
}catch(Exception e){
JOptionPane.showMessageDialog(jpane, "继续 输入(数值) 1，不继续 输入(数值) 0 ！");
}
if(tag==1){
recoverData(thread);
getThread();
}
else{
if( (JOptionPane.YES_NO_OPTION)==JOptionPane.CANCEL_OPTION)System.exit(0);
if((JOptionPane.YES_NO_OPTION)==JOptionPane.YES_OPTION) recoverData(thread);
}
}else{
recoverData(thread);
getThread();
}
}
}
// 安全算法
private boolean check(int thread2) {
boolean[] finish = new boolean[5];
Arrays.fill(finish, false);
int[] work = new int[3];
int[] queue = new int[5];
int q=0;//安全序列下标
for(int i = 0;i<3;i++){
work[i] = available[i];
}
int tT = thread2;
while(tT<5){
for(int R=0;R<3;R++){
if((!(finish[tT]==false))||(!(work[R]>=need[tT][R]))){
tT++;
break;
}else{
if(R==2){
for(int m =0;m<3;m++){
work[m] += allocation[tT][m];
}
for(int s:work){
System.out.print(s+" ");
}
System.out.println("");
finish[tT] = true;
queue[q] = tT;
q++;
tT =0;
}
}
}
}
for(int p =0;p<5;p++){
if(finish[p]==false){
JOptionPane.showMessageDialog(jpane, "安全序列生成失败");
return false;
}
}
JOptionPane.showMessageDialog(jpane, "安全序列："+queue[0]+","+queue[1]+","
+queue[2]+","+queue[3]+","+queue[4]);
return true;
}
private boolean safeState(int thread3){
boolean[] finish = new boolean[5];
Arrays.fill(finish, false);
int[] work = new int[3];
int[] queue = new int[5];
int q=0;//安全序列下标
for(int i = 0;i<3;i++){
work[i] = available[i];
}
int tT = thread3;
while(tT<5){
for(int R=0;R<3;R++){
if((!(finish[tT]==false))||(!(work[R]>=need[tT][R]))){
tT++;
break;
}
else{
if(R==2){
for(int m =0;m<3;m++){
work[m] += allocation[tT][m];
}
finish[tT] = true;
queue[q] = tT;
q++;
tT =0;
}
}
// if((finish[tT]==false)&&(work[R]>=need[tT][R])){
// for(int m =0;m<3;m++){
// work[m] += allocation[tT][m];}
// finish[tT] = true;
// queue[q] = tT;
// q++;
// tT=0;
// }else{
// tT++;
// break;
// }
}
}
for(int p =0;p<5;p++){
if(finish[p]==false){
return false;
}
}
return true;
}
private void recoverData(int thread2) { // 生成失败则重置已分配资源
for(int i=0;i<3;i++){
//重新调整系统资源数
available[i]+=request[i];
//计算各个线程拥有资源
allocation[thread2][i]-=request[i];
//重新计算需求
need[thread2][i]+=request[i];
}
}
private void allocateData(int thread2) { //分配
for(int i=0;i<3;i++){
//重新调整可用系统资源数
available[i]-=request[i];
//计算各个线程分配后拥有资源
allocation[thread2][i]+=request[i];
//重新计算需求
need[thread2][i]-=request[i];
}
}
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