252. Meeting Rooms #60
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oda
reviewed
Mar 19, 2025
| sort(intervals.begin(), intervals.end()); | ||
| for (int i = 0; i < intervals.size() - 1; i++) { | ||
| // 現在の終了時間 > 次の開始時間 で比較 | ||
| if (intervals[i][1] > intervals[i + 1][0]) { |
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これは「要素間」に対してループを回しているイメージですね。
これでもいいですが、私は要素に対して回して、申し送り事項を作るという形にしたほうが自然に感じます。
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| bool canAttendMeetings(vector<vector<int>>& intervals) { | ||
| set<int> used_hours; | ||
| for (const vector<int>& interval : intervals) { | ||
| // intervalごとにstartからendまでを埋めていく |
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これでもいいですが、たとえば、時間が double になったときに使えないなど、色々と制約はありそうですね。
そういったことも考えていただけるとありがたいです。
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@oda
hour++の部分であったり、埋めていく方式だとそもそもデータ量が多くなりすぎて探索できなさ等ですね。
違う型で渡された場合にも問題ないのか、拡張できるのかは意識するようにいたします。
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問題へのリンク
https://leetcode.com/problems/meeting-rooms/description/
問題文(プレミアムの場合)
Given an array of meeting time intervals where intervals[i] = [starti, endi], determine if a person could attend all meetings.
Example 1:
Input: intervals = [[0,30],[5,10],[15,20]]
Output: false
Example 2:
Input: intervals = [[7,10],[2,4]]
Output: true
Constraints:
0 <= intervals.length <= 104
intervals[i].length == 2
0 <= starti < endi <= 106
備考
次に解く問題の予告
フォルダ構成
LeetCodeの問題ごとにフォルダを作成します。
フォルダ内は、step1.cpp、step2.cpp、step3.cppとsort.cppとtwo_keys.cppとmemo.mdとなります。
memo.md内に各ステップで感じたことを追記します。