Question

这对我来说在逻辑上是完全合理的，但是我的输出是数组随机数的最后一个数字。我怀疑变量会不断被放入正在打印的内容中。我们将非常感谢您的见解。

public class HiitClock extends AsyncTask<String, Void, String> {

    private final String LOG_TAG = HiitClock.class.getSimpleName();
    private TextView clock;
    private StopWatch stopWatch;

    public HiitClock(TextView clock, StopWatch sw){
        this.clock = clock;
        this.clockSegment = clockSegment;
        this.roundSegment = roundSegment;
        this.stopWatch = sw;
    }


    @Override
    protected String doInBackground(String... params) {

        stopWatch.start();

        while(stopWatch.getState() == 1 || stopWatch.getState() == 2){
            publishProgress();
        }

        return "done";
    }

    @Override
    protected void onProgressUpdate(Void... values) {
        super.onProgressUpdate(values);

        clock.setText(convertTime());
    }

    @Override
    protected void onPostExecute(String result) {
        // end of timer
       clock.setText(result);
    }

    /*
    * helper method for converting the time
    * */
    private String convertTime(){

        long milli = stopWatch.getTime();
        long sec = milli / 1000 ;
        long min = sec / 60;
        sec = sec % 60;
        milli = milli % 100;

        return String.format("%02d:%02d:%02d", min, sec, milli);
    }
}

Answer 1

试试这个

public static int[] sortArray(int[] preSortArray, int numberOfValues) {
    int temp;
    for(int i=0; i < numberOfValues.length-i; j++) {

      for(int j=1; j < numberOfValues.length-i; j++){
        if(numberOfValues[j-1] > numberOfValues[j]);
           swap = preSortArray[tempNumTwo];
           preSortArray[j-1] = numberOfValues[j];
           preSortArray[tempNum] = swap;
            }
    }
}return preSortArray;
}

Answer 2

首先，您的交换代码是错误的！第二行应该是：

project2

其次，您对冒泡排序的实现也是错误的。在每次迭代中，我们都不会将第i个数字与第j个数字进行比较！我们比较两个连续的数字。以下是您的代码的更正版本：

preSortArray[i] = preSortArray[tempNumTwo];

冒泡排序输出错误

2 个答案: