{ "data": { "question": { "questionId": "2434", "questionFrontendId": "2349", "categoryTitle": "Algorithms", "boundTopicId": 1687675, "title": "Design a Number Container System", "titleSlug": "design-a-number-container-system", "content": "
Design a number container system that can do the following:
\n\nImplement the NumberContainers
class:
NumberContainers()
Initializes the number container system.void change(int index, int number)
Fills the container at index
with the number
. If there is already a number at that index
, replace it.int find(int number)
Returns the smallest index for the given number
, or -1
if there is no index that is filled by number
in the system.\n
Example 1:
\n\n\nInput\n["NumberContainers", "find", "change", "change", "change", "change", "find", "change", "find"]\n[[], [10], [2, 10], [1, 10], [3, 10], [5, 10], [10], [1, 20], [10]]\nOutput\n[null, -1, null, null, null, null, 1, null, 2]\n\nExplanation\nNumberContainers nc = new NumberContainers();\nnc.find(10); // There is no index that is filled with number 10. Therefore, we return -1.\nnc.change(2, 10); // Your container at index 2 will be filled with number 10.\nnc.change(1, 10); // Your container at index 1 will be filled with number 10.\nnc.change(3, 10); // Your container at index 3 will be filled with number 10.\nnc.change(5, 10); // Your container at index 5 will be filled with number 10.\nnc.find(10); // Number 10 is at the indices 1, 2, 3, and 5. Since the smallest index that is filled with 10 is 1, we return 1.\nnc.change(1, 20); // Your container at index 1 will be filled with number 20. Note that index 1 was filled with 10 and then replaced with 20. \nnc.find(10); // Number 10 is at the indices 2, 3, and 5. The smallest index that is filled with 10 is 2. Therefore, we return 2.\n\n\n
\n
Constraints:
\n\n1 <= index, number <= 109
105
calls will be made in total to change
and find
.设计一个数字容器系统,可以实现以下功能:
\n\n请你实现一个 NumberContainers
类:
NumberContainers()
初始化数字容器系统。void change(int index, int number)
在下标 index
处填入 number
。如果该下标 index
处已经有数字了,那么用 number
替换该数字。int find(int number)
返回给定数字 number
在系统中的最小下标。如果系统中没有 number
,那么返回 -1
。\n\n
示例:
\n\n\n输入:\n[\"NumberContainers\", \"find\", \"change\", \"change\", \"change\", \"change\", \"find\", \"change\", \"find\"]\n[[], [10], [2, 10], [1, 10], [3, 10], [5, 10], [10], [1, 20], [10]]\n输出:\n[null, -1, null, null, null, null, 1, null, 2]\n\n解释:\nNumberContainers nc = new NumberContainers();\nnc.find(10); // 没有数字 10 ,所以返回 -1 。\nnc.change(2, 10); // 容器中下标为 2 处填入数字 10 。\nnc.change(1, 10); // 容器中下标为 1 处填入数字 10 。\nnc.change(3, 10); // 容器中下标为 3 处填入数字 10 。\nnc.change(5, 10); // 容器中下标为 5 处填入数字 10 。\nnc.find(10); // 数字 10 所在的下标为 1 ,2 ,3 和 5 。因为最小下标为 1 ,所以返回 1 。\nnc.change(1, 20); // 容器中下标为 1 处填入数字 20 。注意,下标 1 处之前为 10 ,现在被替换为 20 。\nnc.find(10); // 数字 10 所在下标为 2 ,3 和 5 。最小下标为 2 ,所以返回 2 。\n\n\n
\n\n
提示:
\n\n1 <= index, number <= 109
change
和 find
的 总次数 不超过 105
次。\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, 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Python 2.7.12<\\/code><\\/p>\\r\\n\\r\\n
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\\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
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PHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n