[자바JAVA]Sherlock and Squares 해커랭크

Sherlock and Squares

주어진 수의 범위 안의 제곱근(square integer)의 갯수를 구하는 문제이다.
Watson likes to challenge Sherlock’s math ability. He will provide a starting and ending value that describe a range of integers, inclusive of the endpoints. Sherlock must determine the number of square integers within that range.
Note: A square integer is an integer which is the square of an integer, e.g. 1,4,9.16,25

• 입출력예시1

//입력
3 9

//출력
2

• 입출력예시2

//입력
17 24

//출력
0

풀이

public class SherlockandSquares {
	static int squares(int a, int b) {
		//sol1
		int start = (int) Math.sqrt(a);
		int end = (int) Math.sqrt(b);
		
		start = (Math.pow(start, 2) >= a) ? start-1 : start;

		return end - start;	}

	public static void main(String[] args) {
		System.out.println(squares(24, 49) + ", ans: 3");
		System.out.println(squares(3, 9) + ", ans: 2");
		System.out.println(squares(17, 24) + ", ans: 0");
		System.out.println(squares(100, 1000) + ", ans: 22");
		System.out.println(squares(9, 16) + ", ans: 2");
		System.out.println(squares(11, 734) + ", ans: 24");
		System.out.println(squares(228, 919) + ", ans: 15");
		System.out.println(squares(71, 188) + ", ans: 5"); //9 10 11 12 13
		System.out.println(squares(4, 4) + ", ans: 1");
	}

}

한 줄 풀이

Math 적절히 사용하면 한 줄 코드로 문제 해결아 가능하다.

public class SherlockandSquares {
	static int squares(int a, int b) {		
		//sol2. one line code
		return (int) Math.floor(Math.sqrt(b)) - (int) Math.ceil(Math.sqrt(a)) + 1;
	}

	public static void main(String[] args) {
		System.out.println(squares(24, 49) + ", ans: 3");
		System.out.println(squares(3, 9) + ", ans: 2");
		System.out.println(squares(17, 24) + ", ans: 0");
		System.out.println(squares(100, 1000) + ", ans: 22");
		System.out.println(squares(9, 16) + ", ans: 2");
		System.out.println(squares(11, 734) + ", ans: 24");
	}

}

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