给定二进制字符串中1位和2位字符的计数

给定两个特殊字符，第一个字符可以用0表示，第二个字符可以用10或11表示。现在给出一个由几个位表示的字符串。任务是返回它代表的字符数。请注意，给定的字符串始终有效。

例子：

Input: str = “11100”
Output: 3
“11”, “10” and “0” are the required characters.

Input: str = “100”
Output: 2

方法：解决问题的方法是，如果当前字符为0，则表示单个字符为1位，但是如果当前字符为1，则其后的下一位必须包含在由两位组成的字符，如下所示：没有以1开头的单个位字符。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count
// of required characters
int countChars(string str, int n)
{
  
    int i = 0, cnt = 0;
  
    // While there are characters left
    while (i < n) {
  
        // Single bit character
        if (str[i] == '0')
            i++;
  
        // Two-bit character
        else
            i += 2;
  
        // Update the count
        cnt++;
    }
  
    return cnt;
}
  
// Driver code
int main()
{
    string str = "11010";
    int n = str.length();
  
    cout << countChars(str, n);
  
    return 0;
}

Java

// Java implementation of the above approach
  
class GFG {
      
    // Function to return the count 
    // of required characters 
    static int countChars(String str, int n) 
    { 
      
        int i = 0, cnt = 0; 
      
        // While there are characters left 
        while (i < n) { 
      
            // Single bit character 
            if (str.charAt(i) == '0') 
                i += 1; 
      
            // Two-bit character 
            else
                i += 2; 
      
            // Update the count 
            cnt += 1; 
        } 
      
        return cnt; 
    } 
      
    // Driver code 
    public static void main (String[] args) 
    { 
        String str = "11010"; 
        int n = str.length(); 
      
        System.out.println(countChars(str, n)); 
    } 
    // This code is contributed by AnkitRai01
}

Python3

# Python3 implementation of the approach 
  
# Function to return the count 
# of required characters 
def countChars(string, n) :
  
    i = 0; cnt = 0; 
  
    # While there are characters left 
    while (i < n) :
  
        # Single bit character 
        if (string[i] == '0'):
            i += 1; 
  
        # Two-bit character 
        else :
            i += 2; 
  
        # Update the count 
        cnt += 1; 
  
    return cnt; 
  
# Driver code 
if __name__ == "__main__" : 
  
    string = "11010"; 
    n = len(string); 
  
    print(countChars(string, n));
      
# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;
  
class GFG 
{
      
    // Function to return the count 
    // of required characters 
    static int countChars(string str, int n) 
    { 
      
        int i = 0, cnt = 0; 
      
        // While there are characters left 
        while (i < n)
        { 
      
            // Single bit character 
            if (str[i] == '0') 
                i += 1; 
      
            // Two-bit character 
            else
                i += 2; 
      
            // Update the count 
            cnt += 1; 
        } 
      
        return cnt; 
    } 
      
    // Driver code 
    public static void Main () 
    { 
        string str = "11010"; 
        int n = str.Length; 
      
        Console.WriteLine(countChars(str, n)); 
    } 
}
  
// This code is contributed by AnkitRai01

输出：

3

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