计算由相同字符组成的字符串的子字符串数

给定一个字符串。任务是找出由相同字符组成的子串的数量。

例子：

Input: abba
Output: 5
The desired substrings are {a}, {b}, {b}, {a}, {bb}

Input: bbbcbb
Output: 10

方法：已知长度为n的字符串，共有n*(n+1)/2个子字符串。
让我们将结果初始化为 0。遍历字符串并找到相同字符的连续元素的数量（比如说count ）。每当我们找到另一个字符时，将结果增加count*(count+1)/2 ，将count设置为 1，然后从该索引开始重复上述过程。
请记住，对于每个不同的字符，我们想要的子字符串的数量是 1。

下面是上述方法的实现：

C++

// C++ implementation
// of the above approach
#include 
using namespace std;
 
// Function to return the
// number of substrings of
// same characters
void findNumbers(string s)
{
      if (s.empty()) return 0;
    // Size of the string
    int n = s.size();
 
    // Initialize count to 1
    int count = 1;
    int result = 0;
 
    // Initialize left to 0 and
    // right to 1 to traverse the
    // string
    int left = 0;
    int right = 1;
 
    while (right < n) {
 
        // Checking if consecutive
        // characters are same and
        // increment the count
        if (s[left] == s[right]) {
            count++;
        }
 
        // When we encounter a
        // different characters
        else {
 
            // Increment the result
            result += count * (count + 1) / 2;
 
            // To repeat the whole
            // process set left equals
            // right and count variable to 1
            left = right;
            count = 1;
        }
 
        right++;
    }
 
    // Store the final
    // value of result
    result += count * (count + 1) / 2;
 
    cout << result << endl;
}
 
// Driver code
int main()
{
    string s = "bbbcbb";
 
    findNumbers(s);
}

Java

// Java implementation of the approach
class GFG
{
     
    // Function to return the
    // number of substrings of
    // same characters
    static void findNumbers(String s)
    {
        // Size of the string
        int n = s.length();
     
        // Initialize count to 1
        int count = 1;
        int result = 0;
     
        // Initialize left to 0 and
        // right to 1 to traverse the
        // string
        int left = 0;
        int right = 1;
     
        while (right < n)
        {
     
            // Checking if consecutive
            // characters are same and
            // increment the count
            if (s.charAt(left) == s.charAt(right))
            {
                count++;
            }
     
            // When we encounter a
            // different characters
            else
            {
     
                // Increment the result
                result += count * (count + 1) / 2;
     
                // To repeat the whole
                // process set left equals
                // right and count variable to 1
                left = right;
                count = 1;
            }
     
            right++;
        }
     
        // Store the final
        // value of result
        result += count * (count + 1) / 2;
     
        System.out.println(result);
    }
 
    // Driver code
    public static void main (String[] args)
    {
        String s = "bbbcbb";
 
        findNumbers(s);
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the above approach
 
# Function to return the number of
# substrings of same characters
def findNumbers(s):
     
    # Size of the string
    n = len(s)
 
    # Initialize count to 1
    count = 1
    result = 0
 
    # Initialize left to 0 and right to 1
    # to traverse the string
    left = 0
    right = 1
 
    while (right < n):
 
        # Checking if consecutive
        # characters are same and
        # increment the count
        if (s[left] == s[right]):
            count += 1
 
        # When we encounter a
        # different characters
        else:
 
            # Increment the result
            result += count * (count + 1) // 2
 
            # To repeat the whole
            # process set left equals
            # right and count variable to 1
            left = right
            count = 1
 
        right += 1
 
    # Store the final value of result
    result += count * (count + 1) // 2
 
    print(result)
 
# Driver code
s = "bbbcbb"
 
findNumbers(s)
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
    // Function to return the
    // number of substrings of
    // same characters
    static void findNumbers(String s)
    {
        // Size of the string
        int n = s.Length;
 
        // Initialize count to 1
        int count = 1;
        int result = 0;
 
        // Initialize left to 0 and
        // right to 1 to traverse the
        // string
        int left = 0;
        int right = 1;
 
        while (right < n)
        {
            // Checking if consecutive
            // characters are same and
            // increment the count
            if (s[left] == s[right])
                count++;
 
            // When we encounter a
            // different characters
            else
            {
                // Increment the result
                result += count * (count + 1) / 2;
 
                // To repeat the whole
                // process set left equals
                // right and count variable to 1
                left = right;
                count = 1;
            }
            right++;
        }
 
        // Store the final
        // value of result
        result += count * (count + 1) / 2;
 
        Console.WriteLine(result);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String s = "bbbcbb";
 
        findNumbers(s);
    }
}
 
// This code is contributed by
// sanjeev2552

Javascript

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