表示给定字符串所需的棒数

给定一个由大写字母和数字组成的字符串str ，任务是找到表示它所需的火柴棍数量。

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. 这个想法是存储表示特定字母和数字所需的火柴棒计数，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棒计数。

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

Javascript

// Javascipt implementation of the
// above approach
 
// stick[] stores the count
// of matchsticks required to
// represent the alphabets
var 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
var number = [ 6, 2, 5, 5, 4, 5, 6,
                 3, 7, 6 ];
                  
// Function that return the count of
// sticks required to represent
// the given string
function countSticks(str)
{
    var cnt = 0;
   
    // For every char of the given
    // string
    for (var i = 0; str[i]; i++) {
   
        var ch = str[i];
   
        // Add the count of sticks
        // required to represent the
        // current character
        if (ch >= 'A' && ch <= 'Z') {
            cnt += sticks[ch.charCodeAt(0) - 'A'.charCodeAt(0)];
        }
        else {
            cnt += number[ch.charCodeAt(0) - '0'.charCodeAt(0)];
        }
    }
    return cnt;
}
  
// Driver Code
  
// Given array
 
var str = "GEEKSFORGEEKS";
 
// Function Call
document.write(countSticks(str));
 
// This code is contributed by ShubhamSingh10

输出：

66

时间复杂度： O(N)，其中 N 是给定字符串的长度。