{ "data": { "question": { "questionId": "1000041", "questionFrontendId": "面试题 10.10", "categoryTitle": "LCCI", "boundTopicId": 99444, "title": "Rank from Stream LCCI", "titleSlug": "rank-from-stream-lcci", "content": "
Imagine you are reading in a stream of integers. Periodically, you wish to be able to look up the rank of a number x (the number of values less than or equal to x). lmplement the data structures and algorithms to support these operations. That is, implement the method track (int x), which is called when each number is generated, and the method getRankOfNumber(int x), which returns the number of values less than or equal to x.
Note: This problem is slightly different from the original one in the book.
\r\n\r\nExample:
\r\n\r\n\r\nInput:\r\n["StreamRank", "getRankOfNumber", "track", "getRankOfNumber"]\r\n[[], [1], [0], [0]]\r\nOutput:\r\n[null,0,null,1]\r\n\r\n\r\n
Note:
\r\n\r\nx <= 50000track and getRankOfNumber methods are less than or equal to 2000.假设你正在读取一串整数。每隔一段时间,你希望能找出数字 x 的秩(小于或等于 x 的值的个数)。请实现数据结构和算法来支持这些操作,也就是说:
\n\n实现 track(int x) 方法,每读入一个数字都会调用该方法;
实现 getRankOfNumber(int x) 方法,返回小于或等于 x 的值的个数。
注意:本题相对原题稍作改动
\n\n示例:
\n\n输入:\n["StreamRank", "getRankOfNumber", "track", "getRankOfNumber"]\n[[], [1], [0], [0]]\n输出:\n[null,0,null,1]\n\n\n
提示:
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