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Sep 30, 2025
| public int largestRectangleArea(int[] heights) { | ||
| ArrayDeque<ContinuousBar> continuousBars = new ArrayDeque<>(); | ||
| int maxSize = 0; | ||
| int heightsLength = heights.length; |
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私はこれ変数に置かないかもしれません。(特に扱いやすくなっていないので。)まあ、置いてもいいでしょう。
| - 最後にStackに残った要素を計算する | ||
| - このコードしか思いつかなかったが、ACできなかった理由としては | ||
| - addFirstしているため単調増加Stackを保てなかった | ||
| - 落ちるケース[3,6,5,7,4,8,1,0] |
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できればこういうときは最小のケースを書いておきたいです。
[1, 3, 2, 2] であってますか?
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最小ケースは両端が低くて中盤に山がくる [1, 3, 2, 1] です。
| } | ||
| rectangleCandidate.addLast(new RectangleCandidate(start, heights[i])); | ||
| } | ||
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heights の後ろに0を追加すると、後ろのコード不要になりそうですね。(番兵。)
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確認ありがとうございます！
下記で対応しましたmm
面積を算出するコードが重複しており気になっていたのでリファクタできてよかったです
Step4を追加
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https://leetcode.com/problems/largest-rectangle-in-histogram/description/