코딜리티 - BinaryGap(c++, javascript)

요 며칠간 열이 나서 집에 틀어박혀 잠만 잤다.

오늘 좀 괜찮아져서 다시 시작한다.

 

 

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:

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].

 

풀이

10진수 N을 2로 나눈 나머지인 2진수의 각 자리를 구한다.

flag를 통해 첫 1을 발견 시 flag를 true로 바꾼다.

이후 0에 대하여 카운트를 세다가 다음 값이 1 이고 flag가 true라면  cnt의 값을 max값에 갱신해나간다.

int solution(int N) {
    bool flag = false;
    int max = 0;
    int cnt = 0;
    
    while(N!=0){
        int rest = N%2;
        N /= 2;
        
        if(rest==1 && !flag) flag = true;
        
        if(rest==0 && flag) ++cnt;
        
        if(N && flag && N%2 == 1){
            if(cnt > max) max = cnt;
            cnt = 0;
            flag = false;
        }
    }
    
    return max;
}

 

밑에 자바스크립트 코드는 나~~중에 다시 풀었는데 조건문이 더 간결해졌다!ㅎㅎ

function solution(N) {
    let flag = 0;
    let cnt = 0;
    let max = 0;

    while (N) {
        const binary = N % 2;

        if (binary) {
            max = Math.max(max, cnt);
            flag = 1;
            cnt = 0;
        }

        if (flag && !binary) ++cnt;

        N = Math.floor(N / 2);
    }

    return max;
}