Java中如何取得midi音乐的最大和弦数

midi音乐的最大和弦数就是最大复音数，所谓复音就是同一时间点（tick）有多种声音，而这时的声音的种数就是复音数，最大复音数就是一首乐曲中复音数最大的那个时间点的复音数。在midi中声音是使用事件（Event）来记录的，事件有发生时间（tick）和该事件的内容（MidiMessage，即声音）两个属性。所以，复音数在midi里面就是同一个tick的事件的个数，最大复音数就是其最大值。java程序如下：

    public long getMaxChordNum(String midi); throws Exception {
        File mf = new File(midi);;
        Map m = new HashMap();;
        Track[] tracks = MidiSystem.getSequence(mf);.getTracks();;
        for (int i = 0; i < tracks.length; i++); {
            for (int j = 0; j < tracks[i].size();; j++); {
                Long ltick = new Long(tracks[i].get(j);.getTick(););;
                if (ltick.longValue(); == 0L); {
                    continue;
                }
                if (m.containsKey(ltick);); {
                    Long value = (Long); m.get(ltick);;
                    m.put(ltick, new Long(value.longValue(); + 1););;
                } else {
                    m.put(ltick, new Long(1););;
                }
            }
        }

        Set keys = m.keySet();;
        Iterator keyit = keys.iterator();;
        long max = 0;
        while (keyit.hasNext();); {
            Long key = (Long); keyit.next();;
            Long value = (Long); m.get(key);;
            long tmp = value.longValue();;
            if (tmp > max); {
                max = tmp;
            }
        }
        return max;
    }

效率提高版

    public long getMaxChordNum(Track[] tracks); throws Exception {
        long max = 0;
        if (tracks != null && tracks.length > 0); {
            int size = 0;
            for (int i = 0; i < tracks.length; i++); {
                size += tracks[i].size();;
            }
            long[] ticks = new long[size];
            int idx = 0;
            for (int i = 0; i < tracks.length; i++); {
                for (int j = 0; j < tracks[i].size();; j++); {
                    ticks[idx++] = tracks[i].get(j);.getTick();;
                }
            }
            Arrays.sort(ticks);;
            long pretick = 0;
            long precnt = 1;
            for (int i = 0; i < size; i++); {
                if (ticks[i] == 0);
                    continue;
                if (ticks[i] == pretick); {
                    precnt++;
                } else {
                    max = max < precnt ? precnt : max;
                    pretick = ticks[i];
                    precnt = 1;
                }
            }
            max = max < precnt ? precnt : max;
        }
        return max;
    }