计算给定二进制字符串的分数

给定一个二进制字符串str 。对于n-contiguous 1s ，分数更新为score = score + n ² ，对于n-contiguous 0s ，分数更新为score = score - n ² 。任务是找到完整的二进制字符串的分数。
例子：

Input: str = 11011 
Output: 7 
score(“11”) – score(“0”) + score(“11”) = 2² – 1² + 2² = 7
Input: str = 1100011 
Output: -1 
 

方法：为了解决问题，遍历给定的字符串并计算连续的1s和0s的数量。对于每个 n 1的连续块，将n ²添加到当前分数，类似地，对于每个 n 0的连续块，从当前分数中减去n ² 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the score for
// the given binary string
int calcScore(string str)
{
    int score = 0;
    int len = str.length();
 
    // Traverse through string character
    for (int i = 0; i < len;) {
 
        // Initialize current chunk's size
        int chunkSize = 1;
 
        // Get current character
        char currentChar = str[i++];
 
        // Calculate total chunk size
        // of same characters
        while (i < len && str[i] == currentChar) {
            chunkSize++;
            i++;
        }
 
        // Add/subtract pow(chunkSize, 2)
        // depending upon character
        if (currentChar == '1')
            score += pow(chunkSize, 2);
        else
            score -= pow(chunkSize, 2);
    }
 
    // Return the score
    return score;
}
 
// Driver code
int main()
{
    string str = "11011";
    cout << calcScore(str);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
    // Function to return the score for
    // the given binary string
    public static int calcScore(String str)
    {
        int score = 0;
        int len = str.length();
 
        // Traverse through string character
        for (int i = 0; i < len;)
        {
 
            // Initialize current chunk's size
            int chunkSize = 1;
 
            // Get current character
            char currentChar = str.charAt(i++);
 
            // Calculate total chunk size
            // of same characters
            while (i < len && str.charAt(i) == currentChar)
            {
                chunkSize++;
                i++;
            }
 
            // Add/subtract pow(chunkSize, 2)
            // depending upon character
            if (currentChar == '1')
                score += Math.pow(chunkSize, 2);
            else
                score -= Math.pow(chunkSize, 2);
        }
         
        // Return the score
        return score;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str = "11011";
        System.out.println(calcScore(str));
    }
}
 
// This code is contributed by Naman_Garg

Python3

# Python 3 implementation of the approach
 
# Function to return the score for
# the given binary string
def calcScore(str):
    score = 0
    len1 = len(str)
 
    # Traverse through string character
    i = 0
    while(i < len1):
         
        # Initialize current chunk's size
        chunkSize = 1
 
        # Get current character
        currentChar = str[i]
        i += 1
 
        # Calculate total chunk size
        # of same characters
        while (i < len1 and str[i] == currentChar):
            chunkSize += 1
            i += 1
 
        # Add/subtract pow(chunkSize, 2)
        # depending upon character
        if (currentChar == '1'):
            score += pow(chunkSize, 2)
        else:
            score -= pow(chunkSize, 2)
 
    # Return the score
    return score
 
# Driver code
if __name__ == '__main__':
    str = "11011"
    print(calcScore(str))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
     
class GFG
{
     
    // Function to return the score for
    // the given binary string
    public static int calcScore(String str)
    {
        int score = 0;
        int len = str.Length;
 
        // Traverse through string character
        for (int i = 0; i < len;)
        {
 
            // Initialize current chunk's size
            int chunkSize = 1;
 
            // Get current character
            char currentChar = str[i++];
 
            // Calculate total chunk size
            // of same characters
            while (i < len && str[i] == currentChar)
            {
                chunkSize++;
                i++;
            }
 
            // Add/subtract pow(chunkSize, 2)
            // depending upon character
            if (currentChar == '1')
                score += (int)Math.Pow(chunkSize, 2);
            else
                score -= (int)Math.Pow(chunkSize, 2);
        }
         
        // Return the score
        return score;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String str = "11011";
        Console.WriteLine(calcScore(str));
    }
}
 
// This code contributed by Rajput-Ji

Javascript

 
// Javascript implementation of the approach
 
// Function to return the score for
// the given binary string
function calcScore(str)
{
    var score = 0;
    var len = str.length;
 
    // Traverse through string character
    for (var i = 0; i < len;) {
 
        // Initialize current chunk's size
        var chunkSize = 1;
 
        // Get current character
        var currentChar = str[i++];
 
        // Calculate total chunk size
        // of same characters
        while (i < len && str[i] == currentChar) {
            chunkSize++;
            i++;
        }
 
        // Add/subtract pow(chunkSize, 2)
        // depending upon character
        if (currentChar == '1')
            score += Math.pow(chunkSize, 2);
        else
            score -= Math.pow(chunkSize, 2);
    }
 
    // Return the score
    return score;
}
 
// Driver code
var str = "11011";
document.write( calcScore(str));
 
 

输出：

7