面试题 16.02. 单词频率

题目描述

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)查询指定单词在数中出现的频率

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

解法

方法一：哈希表

我们用哈希表 \(cnt\) 统计 \(book\) 中每个单词出现的次数。

调用 get 函数时，我们只需要返回 \(cnt\) 中对应的单词的出现次数即可。

时间复杂度方面，初始化哈希表 \(cnt\) 的时间复杂度为 \(O(n)\)，其中 \(n\) 为 \(book\) 的长度。get 函数的时间复杂度为 \(O(1)\)。空间复杂度为 \(O(n)\)。

Python3JavaC++GoTypeScriptRustJavaScriptSwift

class WordsFrequency:
    def __init__(self, book: List[str]):
        self.cnt = Counter(book)

    def get(self, word: str) -> int:
        return self.cnt[word]


# Your WordsFrequency object will be instantiated and called as such:
# obj = WordsFrequency(book)
# param_1 = obj.get(word)

class WordsFrequency {
    private Map<String, Integer> cnt = new HashMap<>();

    public WordsFrequency(String[] book) {
        for (String x : book) {
            cnt.merge(x, 1, Integer::sum);
        }
    }

    public int get(String word) {
        return cnt.getOrDefault(word, 0);
    }
}

/**
 * Your WordsFrequency object will be instantiated and called as such:
 * WordsFrequency obj = new WordsFrequency(book);
 * int param_1 = obj.get(word);
 */

class WordsFrequency {
public:
    WordsFrequency(vector<string>& book) {
        for (auto& x : book) {
            ++cnt[x];
        }
    }

    int get(string word) {
        return cnt[word];
    }

private:
    unordered_map<string, int> cnt;
};

/**
 * Your WordsFrequency object will be instantiated and called as such:
 * WordsFrequency* obj = new WordsFrequency(book);
 * int param_1 = obj->get(word);
 */

type WordsFrequency struct {
    cnt map[string]int
}

func Constructor(book []string) WordsFrequency {
    cnt := map[string]int{}
    for _, x := range book {
        cnt[x]++
    }
    return WordsFrequency{cnt}
}

func (this *WordsFrequency) Get(word string) int {
    return this.cnt[word]
}

/**
 * Your WordsFrequency object will be instantiated and called as such:
 * obj := Constructor(book);
 * param_1 := obj.Get(word);
 */

class WordsFrequency {
    private cnt: Map<string, number>;

    constructor(book: string[]) {
        const cnt = new Map<string, number>();
        for (const word of book) {
            cnt.set(word, (cnt.get(word) ?? 0) + 1);
        }
        this.cnt = cnt;
    }

    get(word: string): number {
        return this.cnt.get(word) ?? 0;
    }
}

/**
 * Your WordsFrequency object will be instantiated and called as such:
 * var obj = new WordsFrequency(book)
 * var param_1 = obj.get(word)
 */

use std::collections::HashMap;
struct WordsFrequency {
    cnt: HashMap<String, i32>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl WordsFrequency {
    fn new(book: Vec<String>) -> Self {
        let mut cnt = HashMap::new();
        for word in book.into_iter() {
            *cnt.entry(word).or_insert(0) += 1;
        }
        Self { cnt }
    }

    fn get(&self, word: String) -> i32 {
        *self.cnt.get(&word).unwrap_or(&0)
    }
}

/**
 * @param {string[]} book
 */
var WordsFrequency = function (book) {
    this.cnt = new Map();
    for (const x of book) {
        this.cnt.set(x, (this.cnt.get(x) || 0) + 1);
    }
};

/**
 * @param {string} word
 * @return {number}
 */
WordsFrequency.prototype.get = function (word) {
    return this.cnt.get(word) || 0;
};

/**
 * Your WordsFrequency object will be instantiated and called as such:
 * var obj = new WordsFrequency(book)
 * var param_1 = obj.get(word)
 */

class WordsFrequency {
    private var cnt: [String: Int] = [:]

    init(_ book: [String]) {
        for word in book {
            cnt[word, default: 0] += 1
        }
    }

    func get(_ word: String) -> Int {
        return cnt[word, default: 0]
    }
}

面试题 16.02. 单词频率

题目描述

解法

方法一：哈希表

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