[分享] Codility - Binary Gap
灌水part.2
問題是求出正整數的 binary gap。
從問題描述可以看出此平台 int 是 32bits，
正整數的最大值為
0111 1111 1111 1111 1111 1111 1111 1111 = 2,147,483,647
======
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].
===== 以下是滿分解答之一 ===========
int solution(int N) {
// write your code in C++14 (g++ 6.2.0)
int bits[32] = {0}; // 32bits integer initialized by 0
for (int ix = 0; ix < 31; ix++) // bit 31 of unsigned integer is always 0
{
bits[ix] = N % 2;
N = N / 2;
}
// Find start position of 1st 1 appearance.
int start_pos = 30;
for (; start_pos >= 0; start_pos--)
{
if (bits[start_pos] == 0)
continue;
break;
}
int bgap_max = 0, bgap = 0;
for (int ix = start_pos-1; ix >= 0; ix--)
{
if (bits[ix] == 0)
bgap++;
else
{
if (bgap > bgap_max)
{
bgap_max = bgap;
}
bgap = 0;
}
}
return bgap_max;
}
--
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