计数具有相同邻居的字符

给定一个字符串，任务是找到字符与同相邻的字符数。
注意：第一个和最后一个字符将始终被计数，因为它们只有一个相邻的字符。

例子：

Input: str = “egeeksk”
Output: 4
Characters with same adjacent characters are e, g, s, k

Input: str = “eeeeeeee”
Output: 8
Characters with same adjacent characters are e, e, e, e, e, e, e, e

方法：

1. 如果字符串的长度小于3，则返回字符串的长度。
2. 用2初始化计数，因为第一个和最后一个总是被计数。
3. 开始遍历字符串。
   • 检查当前字符的前面和后面的字符是相同的。
   • 增量计数(如果是)。
4. 返回计数。

下面是上述方法的实现：

C++

// C++ implementation of bove approach
#include 
using namespace std;
  
// Function to count the characters
// with same adjacent chracters
int countChar(string str)
{
    int n = str.length();
  
    // if length is less than 3
    // then return length as there
    // will be only two characters
    if (n <= 2)
        return n;
    int count = 2;
  
    // Traverse the string
    for (int i = 1; i < n - 1; i++)
  
        // Increment the count if the previous
        // and next character is same
        if (str[i - 1] == str[i + 1])
            count++;
  
    // Return count
    return count;
}
  
// Driver code
int main()
{
    string str = "egeeksk";
    cout << countChar(str);
  
    return 0;
}

Java

// Java implementation of the above approach
              
class GFG
{
          
         // Function to count the characters
         // with same adjacent chracters
      
        static int countChar(String str)
        {
            int n = str.length();
          
            // if length is less than 3
            // then return length as there
            // will be only two characters
            if (n <= 2)
                return n;
            int count = 2;
          
            // Traverse the string
            for (int i = 1; i < n - 1; i++)
          
                // Increment the count if the previous
                // and next character is same
                if (str.charAt(i - 1) == str.charAt(i + 1))
                    count++;
          
            // Return count
            return count;
        }
      
        // Driver code
        public static void main(String []args)
        {
            String str = "egeeksk";
            System.out.println(countChar(str));
          
          
        }
}
  
  
// This code is contributed
// by ihritik

Python3

# Python 3 implementation of bove approach
  
# Function to count the characters
# with same adjacent chracters
def countChar(str):
    n = len(str)
  
    # if length is less than 3
    # then return length as there
    # will be only two characters
    if (n <= 2):
        return n
    count = 2
  
    # Traverse the string
    for i in range(1, n - 1):
          
        # Increment the count if the previous
        # and next character is same
        if (str[i - 1] == str[i + 1]):
            count += 1
  
    # Return count
    return count
  
# Driver code
if __name__ == '__main__':
    str = "egeeksk"
    print(countChar(str))
      
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of above approach
using System;
  
class GFG
{
  
// Function to count the characters
// with same adjacent chracters
static int countChar(String str)
{
    int n = str.Length;
  
    // if length is less than 3
    // then return length as there
    // will be only two characters
    if (n <= 2)
        return n;
    int count = 2;
  
    // Traverse the string
    for (int i = 1; i < n - 1; i++)
  
        // Increment the count if the previous
        // and next character is same
        if (str[i - 1] == str[i + 1])
            count++;
  
    // Return count
    return count;
}
  
// Driver code
public static void Main()
{
    String str = "egeeksk";
    Console.WriteLine(countChar(str));
}
}
  
// This code is contributed
// by Subhadeep

输出：

4