Question

最近，我借助Tree数据结构实现了电话簿。

我不仅在寻找可行的解决方案，而且还在寻找运行时和存储方面的最佳解决方案。

根据我的计算，需要N*log(T)到add()和get()，其中N是多个入口字符，T是每个节点的子代数目。 / p>

我可以解释一下我的想法：插入一个新名称后，需要花费一些时间才能在子Node中找到一个具有相同字符的TreeMap = log(T)，并且在N次发生参数String中的每个字符。

我对吗？

public final class PhoneBook {

private final Node name;

private final Node surname;

private final Comparator<Record> comparator;

public PhoneBook() {
    this.name = new Node();
    this.surname = new Node();
    comparator = (r1, r2) -> {
        int result = r1.getName().compareTo(r2.getName());
        if (result == 0) {
            result = r1.getSurname().compareTo(r2.getSurname());
            if (result == 0) {
                result = r1.getNumber().compareTo(r2.getNumber());
            }
        }
        return result;
    };
}

public void add(final Record record) {
    add(record.getName().toLowerCase(), record, name);
    add(record.getSurname().toLowerCase(), record, surname);
}

public SortedSet<Record> get(final String prefix) {
    final String lc = prefix.toLowerCase();
    final List<Record> recordsRetrievedByName = get(lc, name);
    final List<Record> recordsRetrievedBySurname = get(lc, surname);
    final SortedSet<Record> result = new TreeSet<>(comparator);
    result.addAll(recordsRetrievedByName);
    result.addAll(recordsRetrievedBySurname);
    return result;
}

private List<Record> get(final String prefix, final Node ancestor) {
    Node node = ancestor;
    for (final char c: prefix.toCharArray()) {
        final Node child = node.children.get(c);
        if (child == null) {
            return Collections.emptyList();
        }
        node = child;
    }
    return node.records;
}

private void add(final String str, final Record record, final Node ancestor) {
    Node node = ancestor;
    for (final char c: str.toCharArray()) {
        final Node child;
        if (node.children.containsKey(c)) {
            child = node.children.get(c);
        } else {
            child = new Node();
            node.children.put(c, child);
        }
        child.records.add(record);
        node = child;
    }
}

private static final class Node {

    private final Map<Character, Node> children = new TreeMap<>();

    private final List<Record> records = new ArrayList<>();

}

}

还有Record不可变对象

public final class Record {

private final String name;

private final String surname;

private final String number;

//constructor, getters, toString
}

您能帮我计算Big-O的复杂度吗？

0 个答案: