实现思路1：时间差

    private long firstTime=0;

    @Override
    public void onBackPressed() {
        long secondTime=System.currentTimeMillis();
        if(secondTime-firstTime>2000){
            Toast.makeText(this,"再按一次退出键退出",Toast.LENGTH_SHORT).show();
            firstTime=secondTime;
        }else{
            super.onBackPressed();
        }
    }```

实现思路2：开启线程，延时处理

private boolean canExit;

private Handler handler=new Handler();

@Override
public void onBackPressed() {
    if (canExit){
        super.onBackPressed();
    }else{
        Toast.makeText(this,"再按一次退出键退出",Toast.LENGTH_SHORT).show();
        canExit=true;
        handler.postDelayed(new Runnable() {
            @Override
            public void run() {
                canExit=false;
            }
        },2000);
    }
}```