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Step3 구현 - #102
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Step3 구현#102
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javajigi
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Oct 12, 2018
| if (this.xPoint == point.xPoint) { | ||
| return true; | ||
| } | ||
| return false; |
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return this.xPoint == point.xPoint;과 같이 구현 가능함.
Y도 같음
| public void toDo(ArrayList<Point> points) { | ||
| HashMap<Integer, String> toDo = new HashMap<>(); | ||
| toDo.put(TWO, new Line(points).calculate()); | ||
| toDo.put(FOUR, new Rectangle(points).calculate()); |
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강의시간에 if 와 switch 를 사용하지말고 구현해 보라고 한걸 듣고 미리 hashMap으로 rectangle과 line 중 선택해 출력하게 구현 해봤습니다.