1. 我的代碼

我的代碼，長得像匹馬，主要是比較時間那一塊寫得太冗長：

    public static class LogSystem {
        //      List<String> stamps = new ArrayList<>();
        HashMap<String[], Integer> map = new HashMap<>();

        public LogSystem() {

        }

        public void put(int id, String timestamp) {
            map.put(timestamp.split(":"), id);
        }

        public List<Integer> retrieve(String s, String e, String gra) {
            int graIndex;
            switch (gra) {
                case "Year":
                    graIndex = 0;
                    break;
                case "Month":
                    graIndex = 1;
                    break;
                case "Day":
                    graIndex = 2;
                    break;
                case "Hour":
                    graIndex = 3;
                    break;
                case "Minute":
                    graIndex = 4;
                    break;
                case "Second":
                    graIndex = 5;
                    break;
                default:
                    graIndex = 0;
            }
            //01這種能parseInt嗎,可以
            String[] sArr = s.split(":");
            String[] eArr = e.split(":");
            ArrayList<Integer> res = new ArrayList<>();
            boolean sTrue = false;
            boolean eTrue = false;
            for (String[] key : map.keySet()) {
                for (int i = 0; i <= graIndex; i++) {
                    if (Integer.parseInt(key[i]) < Integer.parseInt(sArr[i])) {
                        break;
                    }
                    if (i < graIndex && Integer.parseInt(key[i]) > Integer.parseInt(sArr[i])) {
                        sTrue = true;
                        break;
                    }
                    if (i == graIndex && Integer.parseInt(key[i]) >= Integer.parseInt(sArr[i])) {
                        sTrue = true;
                    }
                }
                if (!sTrue){
                    continue;
                }
                for (int i = 0; i <= graIndex; i++) {
                    if (Integer.parseInt(key[i]) > Integer.parseInt(eArr[i])) {
                        break;
                    }

                    if (i < graIndex && Integer.parseInt(key[i]) < Integer.parseInt(eArr[i])) {
                        eTrue = true;
                        break;
                    }
                    if (i == graIndex && Integer.parseInt(key[i]) <= Integer.parseInt(eArr[i])) {
                        eTrue = true;
                    }
                }
                if (eTrue) {
                    res.add(map.get(key));
                }
                sTrue = false;
                eTrue = false;
            }
            return res;
        }
    }

2. 別人簡潔的代碼

首先我摘錄一段String的compareTo方法的comment：

Compares two strings lexicographically.
The comparison is based on the Unicode value of each character in the strings.

所以不需要像我上面那樣，把每個unit分開比較斩松，而只需要比較lexicographically（字典序）就能找出target。

    public static class LogSystem {

        List<String[]> timestamps = new LinkedList<>();
        List<String> units = Arrays.asList("Year", "Month", "Day", "Hour", "Minute", "Second");
        int[] indices = new int[]{4,7,10,13,16,19};

        public void put(int id, String timestamp) {
            timestamps.add(new String[]{Integer.toString(id), timestamp});
        }

        public List<Integer> retrieve(String s, String e, String gra) {
            List<Integer> res = new LinkedList<>();
            int idx = indices[units.indexOf(gra)];
            for (String[] timestamp : timestamps) {
                if (timestamp[1].substring(0, idx).compareTo(s.substring(0, idx)) >= 0 &&
                        timestamp[1].substring(0, idx).compareTo(e.substring(0, idx)) <= 0) {
                    res.add(Integer.parseInt(timestamp[0]));
                }
            }
            return res;
        }
    }