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63 lines
1.7 KiB
Markdown
Executable File
63 lines
1.7 KiB
Markdown
Executable File
# [1154. Day of the Year](https://leetcode.com/problems/day-of-the-year/)
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## 题目
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Given a string `date` representing a [Gregorian calendar](https://en.wikipedia.org/wiki/Gregorian_calendar) date formatted as `YYYY-MM-DD`, return the day number of the year.
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**Example 1:**
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Input: date = "2019-01-09"
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Output: 9
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Explanation: Given date is the 9th day of the year in 2019.
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**Example 2:**
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Input: date = "2019-02-10"
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Output: 41
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**Example 3:**
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Input: date = "2003-03-01"
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Output: 60
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**Example 4:**
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Input: date = "2004-03-01"
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Output: 61
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**Constraints:**
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- `date.length == 10`
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- `date[4] == date[7] == '-'`, and all other `date[i]`'s are digits
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- `date` represents a calendar date between Jan 1st, 1900 and Dec 31, 2019.
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## 题目大意
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实现一个 MajorityChecker 的类，它应该具有下述几个 API：
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- MajorityChecker(int[] arr) 会用给定的数组 arr 来构造一个 MajorityChecker 的实例。
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- int query(int left, int right, int threshold) 有这么几个参数：
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- 0 <= left <= right < arr.length 表示数组 arr 的子数组的长度。
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- 2 * threshold > right - left + 1，也就是说阈值 threshold 始终比子序列长度的一半还要大。
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每次查询 query(...) 会返回在 arr[left], arr[left+1], ..., arr[right] 中至少出现阈值次数 threshold 的元素，如果不存在这样的元素，就返回 -1。
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提示：
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- 1 <= arr.length <= 20000
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- 1 <= arr[i] <= 20000
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- 对于每次查询，0 <= left <= right < len(arr)
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- 对于每次查询，2 * threshold > right - left + 1
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- 查询次数最多为 10000
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## 解题思路
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- 给出一个时间字符串，求出这一天是这一年当中的第几天。
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- 简单题。依照题意处理即可。
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