We have given a string str containing the numeric and alphabetical characters. We need to find the sum of all numbers represented by a continuous sequence of digits available in the given string.
str = “12were43”
55
The sum of 12 and 43 is equal to 55.
str = “1a2c3d”
6
The sum of 1, 2 and 3 is 6.
str = “werderfrewsf”
0
It gives 0 in the output as the string contains no digit.
Our logic to solve the problem is to extract all numbers from the given string and sum them.
In this approach, we will use isDigit() method to check whether the current character is a digit. Also, we multiply the current value of the number by 10 and add the current character to the number if the current character is a digit.
Step 1 - Initialize the 'number' and 'sum' variables to zero.
Step 2 − Iterate through the string and check current character is between 0-9 using the isDigit() method.
Step 3 - If the current character is a number, multiply the numeric value by 10 and add the current numeric value.
Step 4 - If the current character is not a number, add the value of the "number" variable to the "sum" variable and update the value of the "number" variable to zero.
Step 5 − Once the iteration of the loop completes, add the value of the 'number' to the 'sum' variable and return the value of the sum variable.
#include <bits/stdc++.h> using namespace std; // function to return the sum of the consecutive number present in the string int getSumOfDigits(string str){ // store the current number int number = 0; // Stores total sum int sum = 0; // Traverse the string for (auto &ch : str){ // If the current character is between '0' and '9', append it to the number if (isdigit(ch)) { number = number * 10 + ch - '0'; } else { // if the current character is not between '0' and '9', add 'number' to the sum and reset 'number' sum += number; number = 0; } } // if the number is greater than 0, add it to sum sum += number; return sum; } int main(){ string str = "6we24er5rd6"; cout << "The sum of consecutive digits in the given string is - " << getSumOfDigits(str); return 0; }
The sum of consecutive digits in the given string is - 41
Time complexity - O(n), because we only use one loop.
Space Complexity − O(1) since we don’t use any extra space.
In this approach, we use the ASCII values of the character to check whether the current character is a digit. Also, we append characters to the 'number' variable until we get digits in the string and use the atoi() method to extract the number from the string.
Step 1 - Define the 'number' variable and initialize it to an empty string. Also, define the 'sum' variable and initialize it to 0.
Step 2 − Use for loop to traverse the string and get each character of the string.
Step 3 - If c-‘0’ is greater than or equal to zero and less than or equal to 9, it means that the current character is a number.
Step 4 − If the current character is a digit, append it to the ‘number’ string.
Step 5 − If the current character is not a digit, use the c_str() method to convert the number string to a character array and pass it as a parameter of the atoi() method to convert the string to a number. Also, update the number string with the “” value.
The atoi() method returns a number if the string is convertible to a number; Otherwise, it returns zero.
Step 6 − Once the iteration of for loop completes, again use the atoi() method to convert the string to a number and add to the sum value.
#include <bits/stdc++.h> using namespace std; // function to return the sum of the consecutive numbers present in the string int getSumOfDigits(string str){ string number = ""; // to store the sum of all the consecutive numbers int sum = 0; // traverse the string for (char c : str){ // if the current character is between 0 to 9 if (c - '0' >= 0 && c - '0' <= 9){ // append it to the number string number += c; } // if the current character is an alphabet else { // convert string to an array of characters and pass it to atoi() function sum += atoi(number.c_str()); // reset temporary string to empty number = ""; } } // if the number is greater than 0, add it to sum sum += atoi(number.c_str()); return sum; } int main(){ string str = "11aa32bbb5"; cout << "The sum of consecutive digits in the given string is - " << getSumOfDigits(str); return 0; }
The sum of consecutive digits in the given string is - 48
Time complexity - O(N)
Space complexity − O(1)
In this method, we use regular expressions to find matches of all numbers. After that we can convert the string to number and add it to the sum variable.
Step 1 − Define the regex pattern.
Step 2 − Use the regex_search() method to find the match for the number string.
Step 3 − Make iterations using a while loop as long as we find matches.
Step 4 − In the while loop, use the stoi() method to convert the string to a number and add it to the sum variable.
Step 5 - Again, update the string using the match().suffix() method. This way we won't get duplicate matches.
#include <bits/stdc++.h> using namespace std; // Function to calculate the sum of the numbers found in the string int getSumOfDigits(string str){ // regex pattern to find the numbers in the string regex pattern("d+"); smatch match; // variable to store the sum of the numbers int sum = 0; // using the regex_search() function to find the numbers while (regex_search(str, match, pattern)){ // adding the numbers to the sum variable sum += stoi(match[0].str()); // update the string str = match.suffix().str(); } return sum; } int main(){ // input alphanumeric string string str = "abc23@12"; cout << "The sum of consecutive digits in the given string is - " << getSumOfDigits(str); return 0; }
The sum of consecutive digits in the given string is - 0
Time complexity − O(N), as regex finds matches by iterating through the string.
Space complexity − O(1)
We learned three different ways to find the sum of consecutive numbers in a string. The last method is the most optimized code because it uses regular expressions. However, working with regular expressions can be difficult for beginners.
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