面试题集合

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.

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.

Assume that:

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

Complexity:

expected worst-case time complexity is O(log(N));

expected worst-case space complexity is O(1).

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下面是我的方法，复杂度与N中1的个数成正比

int GethighestBit(int n)
{
    if(n < 2)
        return 0;
    else if(n >=2 && n < 4)
        return 1;
    else if(n >=4 && n < 8)
        return 2;
    else if(n >=8 && n < 16)
        return 3;
    else if(n >=16 && n < 32)
        return 4;
    else if(n >=32 && n < 64)
        return 5;
    else if(n >=64 && n < 128)
        return 6;
    else if(n >=128 && n < 256)
        return 7;
    else if(n >=256 && n < 512)
        return 8;
    else if(n >=512 && n < 1024)
        return 9;
    else if(n >=1024 && n < 2048)
        return 10;
    return -1;
}

下面是测试代码
    int n = 1025;
    int maxgap = 0;
    while (n != 0)
    {
        int h1 = GethighestBit(n);
        if(h1 == -1)
        {
            break; //error;
        }
        n -= pow(2, h1);
        int h2 = GethighestBit(n);
        int gap = h1 - h2 -1;
        gap = gap > h1? 0:gap;
        maxgap = gap > maxgap? gap: maxgap;
    }

这个问题的一个小小的陷阱就是10000的gap是不存在的，也任一认为是0，因为其只有一个1，无法形成1和1之间这样的局面。

http://www.ahathinking.com/archives/214.html 给的解答是跟N的二进制表示的位数成正比的，从计算复杂度上看稍逊，但是他的算法不需要计算pow，也不用if else，算法更简练，也是很好的。