{ "data": { "question": { "questionId": "100228", "questionFrontendId": "面试题 03.06", "categoryTitle": "LCCI", "boundTopicId": 58152, "title": "Animal Shelter LCCI", "titleSlug": "animal-shelter-lcci", "content": "
An animal shelter, which holds only dogs and cats, operates on a strictly"first in, first out" basis. People must adopt either the"oldest" (based on arrival time) of all animals at the shelter, or they can select whether they would prefer a dog or a cat (and will receive the oldest animal of that type). They cannot select which specific animal they would like. Create the data structures to maintain this system and implement operations such as enqueue
, dequeueAny
, dequeueDog
, and dequeueCat
. You may use the built-in Linked list data structure.
enqueue
method has a animal
parameter, animal[0]
represents the number of the animal, animal[1]
represents the type of the animal, 0 for cat and 1 for dog.
dequeue*
method returns [animal number, animal type]
, if there's no animal that can be adopted, return [-1, -1]
.
Example1:
\r\n\r\n\r\n Input: \r\n["AnimalShelf", "enqueue", "enqueue", "dequeueCat", "dequeueDog", "dequeueAny"]\r\n[[], [[0, 0]], [[1, 0]], [], [], []]\r\n Output: \r\n[null,null,null,[0,0],[-1,-1],[1,0]]\r\n\r\n\r\n
Example2:
\r\n\r\n\r\n Input: \r\n["AnimalShelf", "enqueue", "enqueue", "enqueue", "dequeueDog", "dequeueCat", "dequeueAny"]\r\n[[], [[0, 0]], [[1, 0]], [[2, 1]], [], [], []]\r\n Output: \r\n[null,null,null,null,[2,1],[0,0],[1,0]]\r\n\r\n\r\n
Note:
\r\n\r\n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:
\n\n输入:\n["AnimalShelf", "enqueue", "enqueue", "dequeueCat", "dequeueDog", "dequeueAny"]\n[[], [[0, 0]], [[1, 0]], [], [], []]\n 输出:\n[null,null,null,[0,0],[-1,-1],[1,0]]\n\n\n
示例2:
\n\n输入:\n["AnimalShelf", "enqueue", "enqueue", "enqueue", "dequeueDog", "dequeueCat", "dequeueAny"]\n[[], [[0, 0]], [[1, 0]], [[2, 1]], [], [], []]\n 输出:\n[null,null,null,null,[2,1],[0,0],[1,0]]\n\n\n
说明:
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out-of-bounds<\\/code>\\u548c
use-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n
OpenJDK 17<\\/code>\\u3002\\u53ef\\u4ee5\\u4f7f\\u7528Java 8\\u7684\\u7279\\u6027\\u4f8b\\u5982\\uff0clambda expressions \\u548c stream API\\u3002<\\/p>\\r\\n\\r\\n
Python 2.7.12<\\/code><\\/p>\\r\\n\\r\\n
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out-of-bounds<\\/code>\\u548c
use-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n
\\r\\nstruct hash_entry {\\r\\n int id; \\/* we'll use this field as the key *\\/\\r\\n char name[10];\\r\\n UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry *s) {\\r\\n HASH_ADD_INT(users, id, s);\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n
\\r\\nstruct hash_entry *find_user(int user_id) {\\r\\n struct hash_entry *s;\\r\\n HASH_FIND_INT(users, &user_id, s);\\r\\n return s;\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n
\\r\\nvoid delete_user(struct hash_entry *user) {\\r\\n HASH_DEL(users, user); \\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\"],\"csharp\":[\"C#\",\"
\\/debug:pdbonly<\\/code>)\\u3002<\\/p>\"],\"javascript\":[\"JavaScript\",\"
Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n
--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f \\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\n
Ruby 3.1<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\n
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rust 1.58.1<\\/code><\\/p>\\r\\n\\r\\n
PHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n