253. Meeting Rooms 2 #45
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oda
reviewed
Mar 14, 2025
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| https://github.com/shining-ai/leetcode/pull/56/files | ||
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| - 会議終わったのに部屋を利用してる人がいたら、後から引っ剥がして新しくstartする人に渡す方法 |
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これは意味づけの仕方次第じゃないですか。
たとえば、ここの会社では、会議が始まるときに受付に鍵を借りに来て、終わるときに受付の郵便箱に返します。
あなたは受付です。会議が始まるたびに、郵便箱の中の鍵を回収して、手持ちの鍵を渡します。
会議室の予約の表が与えられるので、最低いくつ鍵が必要か答えてください。
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なるほど、heapqを受付の箱みたいに一時的に貯める場所として考えれば割と自然ですね(変数名が難しいですが。ended_and_not_startedとか？)
ありがとうございます
| ### Step1 | ||
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| - 「数字が同じ場合、endのマイナスをstartでプラスより先にする」というのが結果としてFalseとTrueの順序で達成できている形になっているが、わかりにくいかもしれない | ||
| - ただわかりやすくする手順が、「コメントを書く」とかしか思いつかない |
nittoco
commented
Mar 24, 2025
| sorted_intervals = self.make_sorted_intervals_by_start(intervals) | ||
| ended_and_using = [] | ||
| for start, end in sorted_intervals: | ||
| if ended_and_using and ended_and_using[0] <= start: |
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改めて見るとちょっとわかりにくい気がする(自分へのコメント)
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問題文: https://neetcode.io/problems/meeting-schedule-ii
問題: Given an array of meeting time interval objects consisting of start and end times [[start_1,end_1],[start_2,end_2],...] (start_i < end_i), find the minimum number of days required to schedule all meetings without any conflicts.
Example 1:
Input: intervals = [(0,40),(5,10),(15,20)]
Output: 2
Explanation:
day1: (0,40)
day2: (5,10),(15,20)
Example 2:
Input: intervals = [(4,9)]
Output: 1
Note:
(0,8),(8,10) is not considered a conflict at 8
Constraints:
0 <= intervals.length <= 500
0 <= intervals[i].start < intervals[i].end <= 1,000,000