java - Java 中最接近 Double 的 int

Question

我有一堆 int 值，例如 10, 5, 4, 2, 1。我有一堆 Double List<Double> 中的值比如 0.65 等等

我如何通过这个列表检查每个 Double 最接近哪个 int 值？

我的代码是针对 Android 音乐应用程序的。这是我最终解决这个问题的方法。

            @Override
            public void onClick(View v) {
                String[] noteNames = new String[9];
                double[] notes = new double[9];
                double crotchet = 60.0 / Long.parseLong(setBPM.getText().toString());
                double quaver = crotchet / 2.0;
                double tripletQuaver = crotchet / 3.0;
                double dottedCrotchet = crotchet + quaver;
                double semiquaver = quaver / 2.0; 
                double dottedQuaver = quaver + semiquaver;
                double minum = crotchet * 2.0;
                double dottedMinum = minum + crotchet;
                double semibreve = minum * 2.0;
                notes[0] = semiquaver;
                notes[1] = tripletQuaver;
                notes[2] = quaver;
                notes[3] = dottedQuaver;
                notes[4] = crotchet;
                notes[5] = dottedCrotchet;
                notes[6] = minum;
                notes[7] = dottedMinum;
                notes[8] = semibreve;
                noteNames[0] = "semiquaver";
                noteNames[1] = "tripletQuaver";
                noteNames[2] = "quaver";
                noteNames[3] = "dottedQuaver";
                noteNames[4] = "crotchet";
                noteNames[5] = "dottedCrotchet";
                noteNames[6] = "minum";
                noteNames[7] = "dottedMinum";
                noteNames[8] = "semibreve";
                for(double l : notes) {
                    System.out.println(l);
                }
                double[] tempCurrentDiff = new double[9];
                double[] currentDiff = new double[9];
                for (Double d : beats) {
                    for (int i = 0; i < 9; i++) {
                        currentDiff[i] = Math.abs(notes[i] - d);
                        tempCurrentDiff[i] = Math.abs(notes[i] - d);
                    }
                    Arrays.sort(tempCurrentDiff);
                    for (int i = 0; i < 9; i++){
                        if (currentDiff[i] == tempCurrentDiff[0]) {
                            System.out.println(noteNames[i]);
                            notesView.append(noteNames[i] + " ");
                            break;
                        }
                    }
                }
          }

Answer 1

使用Math.round(double)函数对列表中的每个元素进行四舍五入。 Math.round(double) 返回参数的最接近 long，并向上舍入 ties(0.5)。但是，如果参数为 NaN，则结果为 0。

所以，

1. 使用 Arrays.sort(a) 对给定的整数数组 int[] a 进行排序(如果尚未排序)
2. 浏览列表中的每个双 x
3. 查找整数key = Math.round(x)
4. 使用Arrays.binarySearch(a, key) ;其中 a 是包含整数数组元素的 int[]。如果数组 a 中不存在该元素，则此函数可能会返回负索引。 负索引:(-(插入点) - 1)。代表我们正在搜索的数组的插入点。

示例代码:

double doub[] = {0.5, 2.6, 3.7, 1.7, 4.6};
  long a[] = {1, 4, 3}; //

  Arrays.sort(a);
  for(double x : doub)
  {
      long key = Math.round(x);
      int index = Arrays.binarySearch(a, key);
      if(index < 0) // element wasn't found
           index = - index -1;
      if(index > a.length -1) // key is greater than the maximum element 
         index = a.length - 1;

      System.out.println("double: "+x+" rounded key: "+key+" value: "+a[index]);

  }