【python3】leetcode 868. Binary Gap（easy）_868. binary gap python3-优快云博客

本文链接：https://blog.youkuaiyun.com/maotianyi941005/article/details/86213464

本文深入探讨了BinaryGap问题，即在给定正整数N的二进制表示中找到两个连续1之间的最大距离。通过具体实例，展示了如何利用Python内置的bin()方法快速转换为二进制，并提供了一种不使用bin()的自定义二进制转换方法。

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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.

其实这道题主要点在考如何转二进制吧？？奈何python大法好。。直接有bin()方法。。不过bin方法转的二进制前面两位是0b表示这是一个二进制串，所以需去掉

class Solution:
    def binaryGap(self, N):
        """
        :type N: int
        :rtype: int
        """
        binN = bin(N)[2:]
        dis = 0
        pre1 = -1
        for i in range(len(binN)):
            if binN[i] == '1':
                if pre1 == -1: pre1 = i
                else:
                    dis = max(dis,i - pre1)
                    pre1 = i
        return dis

思考：如何自己转二进制

cclass Solution:
    def binaryGap(self, N):
        """
        :type N: int
        :rtype: int
        """

        dis = 0
        pre1 = -1

        binN = []
        while(N > 0):
            binN.insert(0,N%2)
            N = int(N/2)
        for i in range(len(binN)):
            if binN[i] == 1:
                if pre1 == -1: pre1 = i
                else:
                    dis = max(dis,i - pre1)
                    pre1 = i    
        return dis