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Codility Lesson 1 - BinaryGap

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BinaryGap

Find longest sequence of zeros in binary representation of an integer.

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:

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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].

Copyright 2009–2019 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

Solution

数字循环除以2，直到0。每次取余，以统计1之间的0的个数。

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class Solution {

    public int solution(int N) {

        int count = 0;
        int maxGap = 0;
        boolean first = true;   // 未找到第一个1时
        
        while (N != 0) {
            if (first) {
                if (N % 2 == 1) {
                    first = false;
                }
            } else {
                if (N % 2 == 0) {
                    count ++;
                } else {
                    // 再次遇到1时，计算之间的gap
                    if (count > maxGap) {
                        maxGap = count;
                    }
                    count = 0;
                }
            }
            // 除以2
            N >>= 1;
        }

        return maxGap;
    }
    
    public static void main(String[] args) {
            Solution solution = new Solution();
            System.out.println("Number 9 (" + Integer.toBinaryString(9) + ") gap: " + solution.solution(9));
            System.out.println("Number 529 (" + Integer.toBinaryString(529) + ") gap: " + solution.solution(529));
            System.out.println("Number 20 (" + Integer.toBinaryString(20) + ") gap: " + solution.solution(20));
            System.out.println("Number 15 (" + Integer.toBinaryString(15) + ") gap: " + solution.solution(15));
            System.out.println("Number 32 (" + Integer.toBinaryString(32) + ") gap: " + solution.solution(32));
        }

}

结果如下：

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Number 9 (1001) gap: 2
Number 529 (1000010001) gap: 4
Number 20 (10100) gap: 1
Number 15 (1111) gap: 0
Number 32 (100000) gap: 0