[Codility] Lesson1. BinaryGap

Description

양의 정수 N이 주어지면 가장 긴 이진 간격의 길이를 반환합니다. N에 이진 간격이 없으면 함수는 0을 반환해야 합니다.

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].

Solution 1. Iteration(반복)

public int solution(int N) {
    int ans = 0;
    String binary = "";
    binary = Integer.toBinaryString(N);
//        while(N>0){
//            binary = (N%2)+binary;
//            N = N/2;
//        }

    int start = 0;
    for(int i =0; i<binary.length(); i++){
        if(binary.charAt(i) == '1'){
            ans = Math.max(i-start-1,ans);
        }
    }
    return ans;
}

주어진 정수를 binary String으로 변환한 뒤 시작점을 기록하고 반복하며 1이 나오는 지점과 시작점까지의 거리의 최대값을 반환해줍니다.

Reference

BinaryGap coding task - Learn to Code - Codility