BinaryGap

🔗 문제 링크

https://app.codility.com/programmers/lessons/1-iterations/binary_gap/start/

❔ 문제 설명

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:

def solution(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:

⚠️ 제한사항

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

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💡 풀이 (언어 : Java)

정수를 이진수로 바꾼 문자열에서 하나씩 확인하며 '1'이 최초로 나온 순간부터 다음 '0'이 나올때마다 개수를 카운트한다. 다시 '1'이 나오면 카운트한 개수를 max와 비교해서 max보다 더 크면 갱신해주고 다시 count를 0으로 초기화해주는 알고리즘이다.

class Solution {
    public int solution(int N) {
        String binInt = Integer.toBinaryString(N);
        int max = 0, count = 0;
        boolean flag = false;
        for (char chr : binInt.toCharArray()) {
            if (flag) {
                if (chr == '0')
                    count++;
            }
            if (chr == '1') {
                flag = true;
                if (count > max)
                    max = count;
                count = 0;
            }
        }
        return max;
    }
}