Leetcode 868. Binary Gap

Task description:

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:

def solution(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].

解题思路：
100100010：
1.找到第一个1； 设置变量 current_gap和max_gap=0

2. 此后，每个1都是binary gap的开始或结束标志
   3.找到下一位，如果是0，那么current_gap 加1.
3. 如果是1， 那么current_gap是离得最近的current_gap，检查max_gap是不是最大的,同时current_gap归零

要注意的地方：

1. tailing zero (本文用了m来保证)
2. 位运算，向右移31位

def binary_gap(N): 
    current_gap = 0  # current step
    max_gap = 0
    m = 32 # m must be big enough, otherwise it get mistakes in the second if.
    for i in range(32):
        # find the first 1
        if(N>>i) &1:
            m = i
            max_gap = max(max_gap,current_gap)  
            current_gap =0 # if the bit is 1, set current_gap 0       
        else:
            if i>m:    # make sure the first 0 is behind 1
                current_gap +=1
    return max_gap