题目

设计一个方法，找出任意指定单词在一本书中的出现频率。

你的实现应该支持如下操作：

• WordsFrequency(book)构造函数，参数为字符串数组构成的一本书
• get(word)查询指定单词在书中出现的频率

示例：

WordsFrequency wordsFrequency = new WordsFrequency({"i", "have", "an", "apple", "he", "have", "a", "pen"});
wordsFrequency.get("you"); //返回0，"you"没有出现过
wordsFrequency.get("have"); //返回2，"have"出现2次
wordsFrequency.get("an"); //返回1
wordsFrequency.get("apple"); //返回1
wordsFrequency.get("pen"); //返回1

提示：

• book[i]中只包含小写字母
• 1 <= book.length <= 100000
• 1 <= book[i].length <= 10
• get函数的调用次数不会超过100000

题解

class WordsFrequency {
    /**
     * 考虑使用字典树实现 只有小写字母固定数组大小
     */
    private final WordsFrequency[] children = new WordsFrequency[26];
    /**
     * 单词命中数量
     */
    private int cnt = 0;

    public WordsFrequency(String[] book) {
        this.init(book);
    }

    private WordsFrequency() {

    }

    protected void init(String[] book) {
        for (String s : book) {
            WordsFrequency it = this;
            for (char c : s.toCharArray()) {
                WordsFrequency child = it.children[c - 'a'];
                // 构建字典树
                it = child == null ? it.children[c - 'a'] = new WordsFrequency() : child;
            }
            it.cnt++;
        }
    }

    public int get(String word) {
        WordsFrequency it = this;
        for (char c : word.toCharArray()) {
            WordsFrequency child = it.children[c - 'a'];
            // 字符命中
            if (child != null) {
                it = child;
            } else {
                return 0;
            }
        }

        return it.cnt;
    }
}