[LeetCode] 868. Binary Gap_Easy

最新推荐文章于 2020-03-01 13:18:13 发布

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最新推荐文章于 2020-03-01 13:18:13 发布

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原文链接：http://www.cnblogs.com/Johnsonxiong/p/9504232.html

本文介绍了一个算法问题：给定一个正整数N，找出并返回N的二进制表示中两个连续1之间的最大距离。若不存在这样的两个连续1，则返回0。文章通过几个示例详细解释了该问题，并提供了一段Python代码实现。

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Given a positive integer N, find and return the longest distance between two consecutive 1's in the binary representation of N.

If there aren't two consecutive 1's, return 0.

Example 1:

Input: 22
Output: 2
Explanation: 
22 in binary is 0b10110.
In the binary representation of 22, there are three ones, and two consecutive pairs of 1's.
The first consecutive pair of 1's have distance 2.
The second consecutive pair of 1's have distance 1.
The answer is the largest of these two distances, which is 2.

Example 2:

Input: 5
Output: 2
Explanation: 
5 in binary is 0b101.

Example 3:

Input: 6
Output: 1
Explanation: 
6 in binary is 0b110.

Example 4:

Input: 8
Output: 0
Explanation: 
8 in binary is 0b1000.
There aren't any consecutive pairs of 1's in the binary representation of 8, so we return 0.

Code

class Solution(object):
    def binaryGap(self, N):
        s, pre, ans = bin(N)[2:], 0, 0
        for index, c in enumerate(s):
            if c == '1':
                ans = max(ans, index - pre)
                pre = index
        return ans