从给定的字符串中找出最大可能的二进制数
给定由集合 {'o '，' n '，' e '，' z '，' r'} 中的字符组成的字符串 str ，任务是通过重新排列给定字符串的字符来找到最大可能的二进制数。请注意，该字符串将至少形成一个有效的数字。
示例:
输入:str = " roeenzooe " 输出:110 “one zero”为必输字符串。
输入:str = " zero zerone " T3】输出: 1000
方法:创建一个地图，并在其中存储【z】和【n】的频率，因为这些是唯一只会出现在 0 或 1 中而不会同时出现在 0 和 1 中的字符。字符串中 1 的数量将等于 'n' 的频率，字符串中 0 的数量将等于地图中 'z' 的频率。现在要找到最大的数字，打印所有的 1，然后是所有的 0。
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 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 required 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 语言(一种计算机语言，尤用于创建网站)
// 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 required 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
Python 3
# 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 required 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 required 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
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return maximum number
// that can be formed from the string
function maxNumber(str, n)
{
// To store the frequency of 'z' and 'n'
// in the given string
var freq = Array(2).fill(0);
for (var 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 required number
var num = "";
// Add all the ones
for (var i = 0; i < freq[1]; i++)
num += '1';
// Add all the zeroes
for (var i = 0; i < freq[0]; i++)
num += '0';
return num;
}
// Driver code
var str = "roenenzooe";
var n = str.length;
document.write( maxNumber(str, n));
</script>
Output:
110
时间复杂度: O(N)
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