代表给定数字所需的火柴计数

给定一个大整数作为字符串str ，任务是找到表示该整数所需的火柴数量。

例子：

Input: str = “56” 
Output: 11 
5 sticks are required to represent 5 and 
6 sticks are required to represent 6.
Input: str = “548712458645878” 
Output: 74 
 

方法：将表示0到9的每个数字所需的火柴计数存储在阵列stick []中。现在逐位遍历给定的字符串，并添加当前数字所需的支数。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// stick[i] stores the count of sticks
// required to represent the digit i
const int sticks[] = { 6, 2, 5, 5, 4, 5,
                       6, 3, 7, 6 };
 
// Function to return the count of
// matchsticks required to represent
// the given number
int countSticks(string str, int n)
{
    int cnt = 0;
 
    // For every digit of the given number
    for (int i = 0; i < n; i++) {
 
        // Add the count of sticks required
        // to represent the current digit
        cnt += (sticks[str[i] - '0']);
    }
 
    return cnt;
}
 
// Driver code
int main()
{
    string str = "56";
    int n = str.length();
 
    cout << countSticks(str, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// stick[i] stores the count of sticks
// required to represent the digit i
static int sticks[] = { 6, 2, 5, 5, 4, 5,
                        6, 3, 7, 6 };
 
// Function to return the count of
// matchsticks required to represent
// the given number
static int countSticks(String str, int n)
{
    int cnt = 0;
 
    // For every digit of the given number
    for (int i = 0; i < n; i++)
    {
 
        // Add the count of sticks required
        // to represent the current digit
        cnt += (sticks[str.charAt(i) - '0']);
    }
    return cnt;
}
 
// Driver code
public static void main(String []args)
{
    String str = "56";
    int n = str.length();
 
    System.out.println(countSticks(str, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# stick[i] stores the count of sticks
# required to represent the digit i
sticks = [ 6, 2, 5, 5, 4, 5,
           6, 3, 7, 6 ];
 
# Function to return the count of
# matchsticks required to represent
# the given number
def countSticks(string, n) :
 
    cnt = 0;
 
    # For every digit of the given number
    for i in range(n) :
 
        # Add the count of sticks required
        # to represent the current digit
        cnt += (sticks[ord(string[i]) - ord('0')]);
 
    return cnt;
 
# Driver code
if __name__ == "__main__" :
 
    string = "56";
    n = len(string);
 
    print(countSticks(string, n));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
                     
class GFG
{
 
// stick[i] stores the count of sticks
// required to represent the digit i
static int []sticks = { 6, 2, 5, 5, 4, 5,
                        6, 3, 7, 6 };
 
// Function to return the count of
// matchsticks required to represent
// the given number
static int countSticks(String str, int n)
{
    int cnt = 0;
 
    // For every digit of the given number
    for (int i = 0; i < n; i++)
    {
 
        // Add the count of sticks required
        // to represent the current digit
        cnt += (sticks[str[i] - '0']);
    }
    return cnt;
}
 
// Driver code
public static void Main(String []args)
{
    String str = "56";
    int n = str.Length;
 
    Console.WriteLine(countSticks(str, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation of the approach
 
// stick[i] stores the count of sticks
// required to represent the digit i
var sticks = [ 6, 2, 5, 5, 4, 5, 6, 3, 7, 6 ]
 
// Function to return the count of
// matchsticks required to represent
// the given number
function countSticks(str, n)
{
    var cnt = 0;
 
    // For every digit of the given number
    for (var i = 0; i < n; i++) {
 
        // Add the count of sticks required
        // to represent the current digit
        cnt += (sticks[str[i] - '0']);
    }
 
    return cnt;
}
 
// Driver code
var str = "56";
var n = str.length;
document.write(countSticks(str, n));
 
// This code is contributed by rutvik_56.

输出：

11

时间复杂度： O(n)