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98784ad3cb
The algorithm I came up with is O(n^2) but given the small numbers of rects we're typically working with, it doesn't really matter. May need to revisit this in the future if we find ourselves with a huge number of rects.
23 lines
488 B
C++
23 lines
488 B
C++
#pragma once
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#include <AK/Vector.h>
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#include <SharedGraphics/Rect.h>
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class DisjointRectSet {
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public:
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DisjointRectSet() { }
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~DisjointRectSet() { }
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DisjointRectSet(DisjointRectSet&& other) : m_rects(move(other.m_rects)) { }
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void add(const Rect&);
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void clear() { m_rects.clear(); }
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void clear_with_capacity() { m_rects.clear_with_capacity(); }
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const Vector<Rect>& rects() const { return m_rects; }
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private:
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void shatter();
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Vector<Rect> m_rects;
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};
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