给定一个二进制数作为长度为L的字符串str 。我们的任务是找到所需的最小操作数，以使该数目变为2 L -1 ，这是一个仅由长度L的1组成的字符串。
在每个操作中，数字N可以替换为N xor(N +1) 。
例子：
Input: str = “10010111”
Output: 5
N = 10010111, N + 1 = 10011000, so N xor (N + 1) = 00001111
N = 00001111, N + 1 = 00010000, so N xor (N + 1) = 00011111
N = 00011111, N + 1 = 00100000, so N xor (N + 1) = 00111111
N = 00111111, N + 1 = 01000000, so N xor (N + 1) = 01111111
N = 01111111, N + 1 = 10000000, so N xor (N + 1) = 11111111
Input: str = “101000100101011101”
Output: 17
方法：执行给定的操作后，可以观察到，为了获得所需的数量，最后，操作数量将为：
Number of Operations = length of the string (after removing leading 0s) – count of consecutive 1’s form the end (starting from the least significant bit)
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the number
// of operations required
int changeToOnes(string str)
{
// ctr will store the number of
// consecutive ones at the end
// of the given binary string
int i, l, ctr = 0;
l = str.length();
// Loop to find number of 1s
// at the end of the string
for (i = l - 1; i >= 0; i--) {
// If the current character is 1
if (str[i] == '1')
ctr++;
// If we encounter the first 0
// from the LSB position then
// we'll break the loop
else
break;
}
// Number of operations
// required is (l - ctr)
return l - ctr;
}
// Function to remove leading
// zeroes from the string
string removeZeroesFromFront(string str)
{
string s;
int i = 0;
// Loop until s[i] becomes
// not equal to 1
while (i < str.length() && str[i] == '0')
i++;
// If we reach the end of
// the string, it means that
// string contains only 0's
if (i == str.length())
s = "0";
// Return the string without
// leading zeros
else
s = str.substr(i, str.length() - i);
return s;
}
// Driver code
int main()
{
string str = "10010111";
// Removing the leading zeroes
str = removeZeroesFromFront(str);
cout << changeToOnes(str);
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function to return the number
// of operations required
static int changeToOnes(String str)
{
// ctr will store the number of
// consecutive ones at the end
// of the given binary string
int i, l, ctr = 0;
l = str.length();
// Loop to find number of 1s
// at the end of the string
for (i = l - 1; i >= 0; i--)
{
// If the current character is 1
if (str.charAt(i) == '1')
ctr++;
// If we encounter the first 0
// from the LSB position then
// we'll break the loop
else
break;
}
// Number of operations
// required is (l - ctr)
return l - ctr;
}
// Function to remove leading
// zeroes from the string
static String removeZeroesFromFront(String str)
{
String s;
int i = 0;
// Loop until s[i] becomes
// not equal to 1
while (i < str.length() &&
str.charAt(i) == '0')
i++;
// If we reach the end of
// the string, it means that
// string contains only 0's
if (i == str.length())
s = "0";
// Return the string without
// leading zeros
else
s = str.substring(i, str.length() - i);
return s;
}
// Driver code
public static void main(String[] args)
{
String str = "10010111";
// Removing the leading zeroes
str = removeZeroesFromFront(str);
System.out.println(changeToOnes(str));
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 implementation of the approach
# Function to return the number
# of operations required
def changeToOnes(string) :
# ctr will store the number of
# consecutive ones at the end
# of the given binary string
ctr = 0;
l = len(string);
# Loop to find number of 1s
# at the end of the string
for i in range(l - 1, -1, -1) :
# If the current character is 1
if (string[i] == '1') :
ctr += 1;
# If we encounter the first 0
# from the LSB position then
# we'll break the loop
else :
break;
# Number of operations
# required is (l - ctr)
return l - ctr;
# Function to remove leading
# zeroes from the string
def removeZeroesFromFront(string) :
s = "";
i = 0;
# Loop until s[i] becomes
# not equal to 1
while (i < len(string) and
string[i] == '0') :
i += 1;
# If we reach the end of
# the string, it means that
# string contains only 0's
if (i == len(string)) :
s = "0";
# Return the string without
# leading zeros
else :
s = string[i: len(string) - i];
return s;
# Driver code
if __name__ == "__main__" :
string = "10010111";
# Removing the leading zeroes
string = removeZeroesFromFront(string);
print(changeToOnes(string));
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the number
// of operations required
static int changeToOnes(String str)
{
// ctr will store the number of
// consecutive ones at the end
// of the given binary string
int i, l, ctr = 0;
l = str.Length;
// Loop to find number of 1s
// at the end of the string
for (i = l - 1; i >= 0; i--)
{
// If the current character is 1
if (str[i] == '1')
ctr++;
// If we encounter the first 0
// from the LSB position then
// we'll break the loop
else
break;
}
// Number of operations
// required is (l - ctr)
return l - ctr;
}
// Function to remove leading
// zeroes from the string
static String removeZeroesFromFront(String str)
{
String s;
int i = 0;
// Loop until s[i] becomes
// not equal to 1
while (i < str.Length &&
str[i] == '0')
i++;
// If we reach the end of
// the string, it means that
// string contains only 0's
if (i == str.Length)
s = "0";
// Return the string without
// leading zeros
else
s = str.Substring(i, str.Length - i);
return s;
}
// Driver code
public static void Main(String[] args)
{
String str = "10010111";
// Removing the leading zeroes
str = removeZeroesFromFront(str);
Console.WriteLine(changeToOnes(str));
}
}
// This code is contributed by 29AjayKumar输出：
5
时间复杂度： O(n)