代表给定字符串所需的曲棍数

给定一个由大写字母和数字组成的字符串str ，任务是找到代表它的火柴棍数量。

例子：

Input: str = “ABC2”
Output: 22
Explanation:
6 sticks are required to represent A,
7 sticks are required to represent B,
4 sticks are required to represent C,
5 sticks are required to represent 2.
Therefore the total number of matchsticks required is 6 + 7 + 4 + 5 = 22.

Input: str = “GEEKSFORGEEKS”
Output: 66
Explanation:
6 sticks are required to represent G,
5 sticks are required to represent E,
4 sticks are required to represent K,
5 sticks are required to represent S,
4 sticks are required to represent F,
6 sticks are required to represent O,
6 sticks are required to represent R.
Therefore the total number of matchsticks required is 6 + 5 + 5 + 4 + 5 + 4 + 6 + 6 + 6 + 5 + 5 + 4 + 5 = 17.

方法：

1. 想法是存储表示特定字母和数字所需的火柴计数，如上图所示。
2. 遍历给定的字符串str并添加每个字符所需的火柴计数。

下面是上述方法的实现：

C++

// C++ implementation of the
// above approach
#include 
using namespace std;
  
// stick[] stores the count
// of matchsticks required to
// represent the alphabets
int sticks[] = { 6, 7, 4, 6, 5, 4, 6,
                 5, 2, 4, 4, 3, 6, 6,
                 6, 5, 7, 6, 5, 3, 5,
                 4, 6, 4, 3, 4 };
  
// number[] stores the count
// of matchsticks required to
// represent the numerals
int number[] = { 6, 2, 5, 5, 4, 5, 6,
                 3, 7, 6 };
  
// Function that return the count of
// sticks required to represent
// the given string
int countSticks(string str)
{
    int cnt = 0;
  
    // For every char of the given
    // string
    for (int i = 0; str[i]; i++) {
  
        char ch = str[i];
  
        // Add the count of sticks
        // required to represent the
        // current character
        if (ch >= 'A' && ch <= 'Z') {
            cnt += sticks[ch - 'A'];
        }
        else {
            cnt += number[ch - '0'];
        }
    }
    return cnt;
}
  
// Driver code
int main()
{
    string str = "GEEKSFORGEEKS";
  
    // Function call to find the
    // count of matchsticks
    cout << countSticks(str);
  
    return 0;
}

Java

// Java implementation of the above approach
class GFG {
      
    // stick[] stores the count
    // of matchsticks required to
    // represent the alphabets
    static int sticks[] = { 6, 7, 4, 6, 5, 4, 6,
                     5, 2, 4, 4, 3, 6, 6,
                     6, 5, 7, 6, 5, 3, 5,
                     4, 6, 4, 3, 4 };
      
    // number[] stores the count
    // of matchsticks required to
    // represent the numerals
    static int number[] = { 6, 2, 5, 5, 4, 5, 6,
                     3, 7, 6 };
      
    // Function that return the count of
    // sticks required to represent
    // the given string
    static int countSticks(String str)
    {
        int cnt = 0;
      
        // For every char of the given
        // string
        for (int i = 0; i < str.length(); i++) {
      
            char ch = str.charAt(i);
      
            // Add the count of sticks
            // required to represent the
            // current character
            if (ch >= 'A' && ch <= 'Z') {
                cnt += sticks[ch - 'A'];
            }
            else {
                cnt += number[ch - '0'];
            }
        }
        return cnt;
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        String str = "GEEKSFORGEEKS";
      
        // Function call to find the
        // count of matchsticks
        System.out.println(countSticks(str));
      
    }
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the above approach 
  
# stick[] stores the count 
# of matchsticks required to 
# represent the alphabets 
sticks = [ 6, 7, 4, 6, 5, 4, 6, 
            5, 2, 4, 4, 3, 6, 6, 
            6, 5, 7, 6, 5, 3, 5, 
            4, 6, 4, 3, 4 ]; 
  
# number[] stores the count 
# of matchsticks required to 
# represent the numerals 
number = [ 6, 2, 5, 5, 4, 5, 6, 3, 7, 6 ]; 
  
# Function that return the count of 
# sticks required to represent 
# the given string 
def countSticks(string) :
  
    cnt = 0; 
  
    # For every char of the given 
    # string 
    for i in range(len(string)) :
  
        ch = string[i]; 
  
        # Add the count of sticks 
        # required to represent the 
        # current character 
        if (ch >= 'A' and ch <= 'Z') :
            cnt += sticks[ord(ch) - ord('A')]; 
          
        else :
            cnt += number[ord(ch) - ord('0')]; 
      
    return cnt; 
  
# Driver code 
if __name__ == "__main__" : 
  
    string = "GEEKSFORGEEKS"; 
  
    # Function call to find the 
    # count of matchsticks 
    print(countSticks(string)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;
  
class GFG
{
      
    // stick[] stores the count
    // of matchsticks required to
    // represent the alphabets
    static int []sticks = { 6, 7, 4, 6, 5, 4, 6,
                    5, 2, 4, 4, 3, 6, 6,
                    6, 5, 7, 6, 5, 3, 5,
                    4, 6, 4, 3, 4 };
      
    // number[] stores the count
    // of matchsticks required to
    // represent the numerals
    static int []number = { 6, 2, 5, 5, 4, 5, 6,
                    3, 7, 6 };
      
    // Function that return the count of
    // sticks required to represent
    // the given string
    static int countSticks(string str)
    {
        int cnt = 0;
      
        // For every char of the given
        // string
        for (int i = 0; i < str.Length; i++) 
        {
      
            char ch = str[i];
      
            // Add the count of sticks
            // required to represent the
            // current character
            if (ch >= 'A' && ch <= 'Z')
            {
                cnt += sticks[ch - 'A'];
            }
            else
            {
                cnt += number[ch - '0'];
            }
        }
        return cnt;
    }
      
    // Driver code
    public static void Main() 
    {
        string str = "GEEKSFORGEEKS";
      
        // Function call to find the
        // count of matchsticks
        Console.WriteLine(countSticks(str));
    }
}
  
// This code is contributed by AnkitRai01

输出：

66

时间复杂度： O(N)，其中N是给定字符串的长度。