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面试题 03.06. 动物收容所

English Version

题目描述

动物收容所。有家动物收容所只收容狗与猫，且严格遵守“先进先出”的原则。在收养该收容所的动物时，收养人只能收养所有动物中“最老”（由其进入收容所的时间长短而定）的动物，或者可以挑选猫或狗（同时必须收养此类动物中“最老”的）。换言之，收养人不能自由挑选想收养的对象。请创建适用于这个系统的数据结构，实现各种操作方法，比如enqueue、dequeueAny、dequeueDog和dequeueCat。允许使用Java内置的LinkedList数据结构。

enqueue方法有一个animal参数，animal[0]代表动物编号，animal[1]代表动物种类，其中 0 代表猫，1 代表狗。

dequeue*方法返回一个列表[动物编号, 动物种类]，若没有可以收养的动物，则返回[-1,-1]。

示例1:

 输入：
["AnimalShelf", "enqueue", "enqueue", "dequeueCat", "dequeueDog", "dequeueAny"]
[[], [[0, 0]], [[1, 0]], [], [], []]
 输出：
[null,null,null,[0,0],[-1,-1],[1,0]]

示例2:

 输入：
["AnimalShelf", "enqueue", "enqueue", "enqueue", "dequeueDog", "dequeueCat", "dequeueAny"]
[[], [[0, 0]], [[1, 0]], [[2, 1]], [], [], []]
 输出：
[null,null,null,null,[2,1],[0,0],[1,0]]

说明:

  1. 收纳所的最大容量为20000

解法

双队列存储。

Python3

class AnimalShelf:
    def __init__(self):
        self.cats = []
        self.dogs = []

    def enqueue(self, animal: List[int]) -> None:
        if animal[1] == 0:
            self.cats.insert(0, animal[0])
        else:
            self.dogs.insert(0, animal[0])

    def dequeueAny(self) -> List[int]:
        if len(self.dogs) == 0:
            return self.dequeueCat()
        if len(self.cats) == 0:
            return self.dequeueDog()
        return self.dequeueDog() if self.dogs[-1] < self.cats[-1] else self.dequeueCat()

    def dequeueDog(self) -> List[int]:
        return [-1, -1] if len(self.dogs) == 0 else [self.dogs.pop(), 1]

    def dequeueCat(self) -> List[int]:
        return [-1, -1] if len(self.cats) == 0 else [self.cats.pop(), 0]


# Your AnimalShelf object will be instantiated and called as such:
# obj = AnimalShelf()
# obj.enqueue(animal)
# param_2 = obj.dequeueAny()
# param_3 = obj.dequeueDog()
# param_4 = obj.dequeueCat()

Java

class AnimalShelf {
    Queue<Integer> cats;
    Queue<Integer> dogs;
    public AnimalShelf() {
        cats = new LinkedList<>();
        dogs = new LinkedList<>();
    }

    public void enqueue(int[] animal) {
        if (animal[1] == 0) {
            cats.offer(animal[0]);
        } else {
            dogs.offer(animal[0]);
        }
    }

    public int[] dequeueAny() {
        return dogs.isEmpty()
            ? dequeueCat()
            : (cats.isEmpty() ? dequeueDog()
                              : (dogs.peek() < cats.peek() ? dequeueDog() : dequeueCat()));
    }

    public int[] dequeueDog() {
        return dogs.isEmpty() ? new int[] {-1, -1} : new int[] {dogs.poll(), 1};
    }

    public int[] dequeueCat() {
        return cats.isEmpty() ? new int[] {-1, -1} : new int[] {cats.poll(), 0};
    }
}

/**
 * Your AnimalShelf object will be instantiated and called as such:
 * AnimalShelf obj = new AnimalShelf();
 * obj.enqueue(animal);
 * int[] param_2 = obj.dequeueAny();
 * int[] param_3 = obj.dequeueDog();
 * int[] param_4 = obj.dequeueCat();
 */

TypeScript

class AnimalShelf {
    private cats: number[];
    private dogs: number[];

    constructor() {
        this.cats = [];
        this.dogs = [];
    }

    enqueue(animal: number[]): void {
        const [i, j] = animal;
        this[j === 0 ? 'cats' : 'dogs'].push(i);
    }

    dequeueAny(): number[] {
        const n = this.dogs.length;
        const m = this.cats.length;
        if (n === 0 && m === 0) {
            return [-1, -1];
        }
        if ((this.dogs[0] ?? Infinity) < (this.cats[0] ?? Infinity)) {
            return [this.dogs.shift(), 1];
        } else {
            return [this.cats.shift(), 0];
        }
    }

    dequeueDog(): number[] {
        if (this.dogs.length === 0) {
            return [-1, -1];
        }
        return [this.dogs.shift(), 1];
    }

    dequeueCat(): number[] {
        if (this.cats.length === 0) {
            return [-1, -1];
        }
        return [this.cats.shift(), 0];
    }
}

/**
 * Your AnimalShelf object will be instantiated and called as such:
 * var obj = new AnimalShelf()
 * obj.enqueue(animal)
 * var param_2 = obj.dequeueAny()
 * var param_3 = obj.dequeueDog()
 * var param_4 = obj.dequeueCat()
 */

Rust

use std::collections::VecDeque;

struct AnimalShelf {
    cats: VecDeque<i32>,
    dogs: VecDeque<i32>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl AnimalShelf {
    fn new() -> Self {
        Self {
            cats: VecDeque::new(),
            dogs: VecDeque::new(),
        }
    }

    fn enqueue(&mut self, animal: Vec<i32>) {
        if animal[1] == 0 {
            self.cats.push_back(animal[0]);
        } else {
            self.dogs.push_back(animal[0]);
        }
    }

    fn dequeue_any(&mut self) -> Vec<i32> {
        match (self.cats.is_empty(), self.dogs.is_empty()) {
            (true, true) => vec![-1, -1],
            (true, false) => self.dequeue_dog(),
            (false, true) => self.dequeue_cat(),
            (false, false) => {
                if self.dogs[0] < self.cats[0] {
                    self.dequeue_dog()
                } else {
                    self.dequeue_cat()
                }
            }
        }
    }

    fn dequeue_dog(&mut self) -> Vec<i32> {
        if self.dogs.is_empty() {
            return vec![-1, -1];
        }
        vec![self.dogs.pop_front().unwrap(), 1]
    }

    fn dequeue_cat(&mut self) -> Vec<i32> {
        if self.cats.is_empty() {
            return vec![-1, -1];
        }
        vec![self.cats.pop_front().unwrap(), 0]
    }
}

/**
 * Your AnimalShelf object will be instantiated and called as such:
 * let obj = AnimalShelf::new();
 * obj.enqueue(animal);
 * let ret_2: Vec<i32> = obj.dequeue_any();
 * let ret_3: Vec<i32> = obj.dequeue_dog();
 * let ret_4: Vec<i32> = obj.dequeue_cat();
 */

...