Convert a String to an Integer using Recursion

Given a string str representing a string, the task is to convert the given string into an integer.

Examples:

Input: str = “1234”
Output: 1234



Input: str = “0145”
Output: 145

Approach: Write a recursive function that will take the first digit of the string and multiply it with the appropriate power of 10 and then add the recursive result for the substring starting at the second index. The termination condition will be when the passed string consists of a single digit. In that case, return the digit represented by the string.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
  
// Recursive function to convert
// string to integer
int stringToInt(string str)
{
  
    // If the number represented as a string
    // contains only a single digit
    // then returns its value
    if (str.length() == 1)
        return (str[0] - '0');
  
    // Recursive call for the sub-string
    // starting at the second character
    double y = stringToInt(str.substr(1));
  
    // First digit of the number
    double x = str[0] - '0';
  
    // First digit multiplied by the
    // appropriate power of 10 and then
    // add the recursive result
    // For example, xy = ((x * 10) + y)
    x = x * pow(10, str.length() - 1) + y;
    return int(x);
}
  
// Driver code
int main()
{
    string str = "1235";
    cout << (stringToInt(str)) << endl;
}
  
// This code is contributed by
// Surendra_Gangwar

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Java

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// Java implementation of the approach
public class GFG {
  
    // Recursive function to convert string to integer
    static int stringToInt(String str)
    {
  
        // If the number represented as a string
        // contains only a single digit
        // then returns its value
        if (str.length() == 1)
            return (str.charAt(0) - '0');
  
        // Recursive call for the sub-string
        // starting at the second character
        double y = stringToInt(str.substring(1));
  
        // First digit of the number
        double x = str.charAt(0) - '0';
  
        // First digit multiplied by the
        // appropriate power of 10 and then
        // add the recursive result
        // For example, xy = ((x * 10) + y)
        x = x * Math.pow(10, str.length() - 1) + y;
        return (int)(x);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String str = "1235";
        System.out.print(stringToInt(str));
    }
}

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Python3

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# Python3 implementation of the approach
  
# Recursive function to convert
# string to integer
def stringToInt(str):
  
    # If the number represented as a string
    # contains only a single digit
    # then returns its value
    if (len(str) == 1):
        return ord(str[0]) - ord('0');
  
    # Recursive call for the sub-string
    # starting at the second character
    y = int(str[1:]);
  
    # First digit of the number
    x = ord(str[0]) - ord('0');
      
    # First digit multiplied by the
    # appropriate power of 10 and then
    # add the recursive result
    # For example, xy = ((x * 10) + y)
    x = x * (10**(len(str) - 1)) + y;
    return int(x);
  
# Driver code
if __name__ == '__main__':
    str = "1235";
    print(stringToInt(str));
  
# This code is contributed by PrinciRaj1992 

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
    // Recursive function to convert string to integer
    static int stringToInt(String str)
    {
  
        // If the number represented as a string
        // contains only a single digit
        // then returns its value
        if (str.Length == 1)
            return (str[0] - '0');
  
        // Recursive call for the sub-string
        // starting at the second character
        double y = stringToInt(str.Substring(1));
  
        // First digit of the number
        double x = str[0] - '0';
  
        // First digit multiplied by the
        // appropriate power of 10 and then
        // add the recursive result
        // For example, xy = ((x * 10) + y)
        x = x * Math.Pow(10, str.Length - 1) + y;
        return (int)(x);
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        String str = "1235";
        Console.Write(stringToInt(str));
    }
}
  
// This code is contributed by Princi Singh

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Output:

1235


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