给定一个字符串str ,该字符串str由集合{‘o’,’n’,’e’,’z’,’r’}中的字符组成,任务是查找可以通过重新排列而形成的最大二进制数给定字符串中的字符。请注意,该字符串将至少形成一个有效数字。
例子:
Input: str = “roenenzooe”
Output: 110
“oneonezero” is the required string
Input: str = “zerozerozeroone”
Output: 1000
方法:创建一个映射并在其中存储“ z”和“ n”的频率,因为这是仅有的字符,它们只会以0或1出现,而不会同时出现在两个字符中。字符串的1的数量将等于“ n”的频率,而字符串的零的数量将等于映射中的“ z”的频率。现在要找到最大的数字,请打印所有数字,然后打印所有零。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return maximum number
// that can be formed from the string
string maxNumber(string str, int n)
{
// To store the frequency of 'z' and 'n'
// in the given string
int freq[2] = { 0 };
for (int i = 0; i < n; i++) {
if (str[i] == 'z') {
// Number of zeroes
freq[0]++;
}
else if (str[i] == 'n') {
// Number of ones
freq[1]++;
}
}
// To store the requried number
string num = "";
// Add all the ones
for (int i = 0; i < freq[1]; i++)
num += '1';
// Add all the zeroes
for (int i = 0; i < freq[0]; i++)
num += '0';
return num;
}
// Driver code
int main()
{
string str = "roenenzooe";
int n = str.length();
cout << maxNumber(str, n);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return maximum number
// that can be formed from the string
static String maxNumber(String str, int n)
{
// To store the frequency of 'z' and 'n'
// in the given string
int[] freq = new int[2];
for (int i = 0; i < n; i++)
{
if (str.charAt(i) == 'z')
// Number of zeroes
freq[0]++;
else if (str.charAt(i) == 'n')
// Number of ones
freq[1]++;
}
// To store the requried number
String num = "";
// Add all the ones
for (int i = 0; i < freq[1]; i++)
num += '1';
// Add all the zeroes
for (int i = 0; i < freq[0]; i++)
num += '0';
return num;
}
// Driver Code
public static void main(String[] args)
{
String str = "roenenzooe";
int n = str.length();
System.out.println(maxNumber(str, n));
}
}
// This code is contributed by
// sanjeev2552
Python3
# Python3 implementation of the approach
# Function to return maximum number
# that can be formed from the string
def maxNumber(string , n) :
# To store the frequency of 'z' and 'n'
# in the given string
freq = [0, 0]
for i in range(n) :
if (string[i] == 'z') :
# Number of zeroes
freq[0] += 1;
elif (string[i] == 'n') :
# Number of ones
freq[1] += 1;
# To store the requried number
num = "";
# Add all the ones
for i in range(freq[1]) :
num += '1';
# Add all the zeroes
for i in range(freq[0]) :
num += '0';
return num;
# Driver code
if __name__ == "__main__" :
string = "roenenzooe";
n = len(string);
print(maxNumber(string, n));
# This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return maximum number
// that can be formed from the string
static string maxNumber(string str, int n)
{
// To store the frequency of 'z' and 'n'
// in the given string
int [] freq = new int[2];
for (int i = 0; i < n; i++)
{
if (str[i] == 'z')
{
// Number of zeroes
freq[0]++;
}
else if (str[i] == 'n')
{
// Number of ones
freq[1]++;
}
}
// To store the requried number
string num = "";
// Add all the ones
for (int i = 0; i < freq[1]; i++)
num += '1';
// Add all the zeroes
for (int i = 0; i < freq[0]; i++)
num += '0';
return num;
}
// Driver code
public static void Main()
{
string str = "roenenzooe";
int n = str.Length;
Console.Write(maxNumber(str, n));
}
}
// This code is contributed by Sanjit Prasad
输出:
110
时间复杂度: O(N)