除以二进制数的 2 的最高幂
给定二进制字符串 str ，任务是求 2 的最大幂，除以给定二进制数的十进制等效值。
示例:
输入: str = "100100" 输出: 2 2 2 = 4 是 2 除以 36 (100100)的最高幂。 输入: str = "10010" 输出: 1
逼近:从右边开始，计算二进制表示中的 0 的个数，这是 2 的最高幂，它将除以这个数。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the highest power of 2
// which divides the given binary number
int highestPower(string str, int len)
{
// To store the highest required power of 2
int ans = 0;
// Counting number of consecutive zeros
// from the end in the given binary string
for (int i = len - 1; i >= 0; i--) {
if (str[i] == '0')
ans++;
else
break;
}
return ans;
}
// Driver code
int main()
{
string str = "100100";
int len = str.length();
cout << highestPower(str, len);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to return the highest power of 2
// which divides the given binary number
static int highestPower(String str, int len)
{
// To store the highest required power of 2
int ans = 0;
// Counting number of consecutive zeros
// from the end in the given binary string
for (int i = len - 1; i >= 0; i--)
{
if (str.charAt(i) == '0')
ans++;
else
break;
}
return ans;
}
// Driver code
public static void main(String[] args)
{
String str = "100100";
int len = str.length();
System.out.println(highestPower(str, len));
}
}
// This code is contributed by Code_Mech.
Python 3
# Python3 implementation of the approach
# Function to return the highest power of 2
# which divides the given binary number
def highestPower(str, length):
# To store the highest required power of 2
ans = 0;
# Counting number of consecutive zeros
# from the end in the given binary string
for i in range(length-1,-1,-1):
if (str[i] == '0'):
ans+=1;
else:
break;
return ans;
# Driver code
def main():
str = "100100";
length = len(str);
print(highestPower(str, length));
if __name__ == '__main__':
main()
# This code contributed by PrinciRaj1992
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the highest power of 2
// which divides the given binary number
static int highestPower(String str, int len)
{
// To store the highest required power of 2
int ans = 0;
// Counting number of consecutive zeros
// from the end in the given binary string
for (int i = len - 1; i >= 0; i--)
{
if (str[i] == '0')
ans++;
else
break;
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
String str = "100100";
int len = str.Length;
Console.WriteLine(highestPower(str, len));
}
}
/* This code contributed by PrinciRaj1992 */
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the highest power of 2
// which divides the given binary number
function highestPower($str, $len)
{
// To store the highest required power of 2
$ans = 0;
// Counting number of consecutive zeros
// from the end in the given binary string
for ($i = $len - 1; $i >= 0; $i--)
{
if ($str[$i] == '0')
$ans++;
else
break;
}
return $ans;
}
// Driver code
$str = "100100";
$len = strlen($str);
echo highestPower($str, $len);
// This code is contributed by Ryuga
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the highest power of 2
// which divides the given binary number
function highestPower(str, len)
{
// To store the highest required power of 2
let ans = 0;
// Counting number of consecutive zeros
// from the end in the given binary string
for (let i = len - 1; i >= 0; i--) {
if (str[i] == '0')
ans++;
else
break;
}
return ans;
}
// Driver code
let str = "100100";
let len = str.length;
document.write(highestPower(str, len));
</script>
Output
2
时间复杂度: O(N)
辅助空间: O(1)
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