[Codillity] Binary Gap

문제

A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.

For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary representation 10100 and contains one binary gap of length 1. The number 15 has binary representation 1111 and has no binary gaps. The number 32 has binary representation 100000 and has no binary gaps.

Write a function:

class Solution { public int solution(int N); }

that, given a positive integer N, returns the length of its longest binary gap. The function should return 0 if N doesn't contain a binary gap.

For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5. Given N = 32 the function should return 0, because N has binary representation '100000' and thus no binary gaps.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..2,147,483,647].
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풀이

1. 해석

10진수를 2진수로 변환했을 때 1과 1사이 존재하는 0의 갯수를 구하라.

이 문제에서 예시로 든 것은 이러하다

• 9를 2진수로 바꾸면 1001 -> 2개
• 1041을 2진수로 바꾸면 1000010001 -> 5개
• 32를 2진수로 바꾸면 10000 -> 0개

2. 코드

// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int N) {
        // write your code in Java SE 11
        String binaryString = Integer.toBinaryString(N);
        char[] binaryArray = binaryString.toCharArray();
        
        int gap = 0;
        int cnt = 0;
        
        for(int i = 0; i < binaryArray.length; i++) {
            if(binaryArray[i] == '1') {
                if(gap > cnt) {
                    cnt = gap;
                }
                gap = 0;
            }
            else {
                    gap++;
            }
        }
        return cnt;
    }
}

• Java에는 Integer 클래스의 Integer.toBinaryString(int N) 함수를 사용하면 2진수로 변환을 해준다.
• 이것을 toCharArray를 이용해 새로운 문자 배열로 변환을 시켜줍니다.
• 이후 1과 1사이에 존재하는 0을 세어줄 gap과 다 세었을때 그 값을 저장할 cnt 를 선언을 해주게 됩니다.
• 문자열을 돌며 값이 1일때 gap, cnt를 비교해 gap이 cnt보다 클 경우 cnt에 gap 값을 넣어주고 0으로 초기화 시켜줍니다.
• 이렇게 문자열 끝까지 반복한 후 cnt 값을 반환해 줍니다.