使用递归将字符串转换为整数

给定一个表示字符串的字符串str ，任务是将给定的字符串转换为整数。

例子：

Input: str = “1234”
Output: 1234

Input: str = “0145”
Output: 145

方法：编写一个递归函数，该函数将字符串的第一位数字乘以适当的10的幂，然后从第二个索引处开始为子字符串添加递归结果。终止条件为传递的字符串包含一位数字时。在这种情况下，返回由字符串表示的数字。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include
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

Java

// 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));
    }
}

Python3

# 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 = stringToInt(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

C#

// 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

输出：

1235