Recursive Implementation of atoi()
The atoi() function takes a string (which represents an integer) as an argument and returns its value. We have discussed iterative implementation of atoi(). How to compute recursively?
Approach:
The idea is to separate the last digit, recursively compute the result for the remaining n-1 digits, multiply the result by 10 and add the obtained value to the last digit.
Below is the implementation of the idea.
C++
#include <cctype>
#include <cstring>
#include <iostream>
using namespace std;
int myAtoiRecursive( char * str, int n)
{
int count = 0, check;
for ( int i = 0; i <= strlen (str); ++i) {
check = isalpha (str[i]);
if (check)
++count;
}
if (count != 0) {
return 0;
}
if (n == 1)
return *str - '0' ;
return (10 * myAtoiRecursive(str, n - 1) + str[n - 1]
- '0' );
}
int main( void )
{
char str[] = "112" ;
int n = strlen (str);
printf ( "%d" , myAtoiRecursive(str, n));
return 0;
}
|
Java
class GFG{
static int myAtoiRecursive(String str, int n)
{
if (str == "" || !str.matches( "^\\d*$" )) {
return 0 ;
}
if (n == 1 )
{
return str.charAt( 0 ) - '0' ;
}
return ( 10 * myAtoiRecursive(str, n - 1 ) +
str.charAt(n - 1 ) - '0' );
}
public static void main(String[] s)
{
String str = "112" ;
int n = str.length();
System.out.println(myAtoiRecursive(str, n));
}
}
|
Python3
def myAtoiRecursive(string, num):
if string.isalpha() :
return 0 ;
if ( len (string) = = 0 ):
return 0 ;
if len (string) = = 1 :
return int (string) + (num * 10 )
num = int (string[ 0 : 1 ]) + (num * 10 )
return myAtoiRecursive(string[ 1 :], num)
string = "112"
print (myAtoiRecursive(string, 0 ))
|
C#
using System;
using System.Text.RegularExpressions;
class GFG{
static int myAtoiRecursive( string str, int n)
{
if (Regex.IsMatch(str, "^[a-zA-Z]*$" )){
return 0;
}
if (n == 1)
{
return str[0] - '0' ;
}
return (10 * myAtoiRecursive(str, n - 1) +
str[n - 1] - '0' );
}
public static void Main()
{
string str = "112" ;
int n = str.Length;
Console.Write(myAtoiRecursive(str, n));
}
}
|
Javascript
<script>
function myAtoiRecursive(str, n)
{
if (str.match(/^[A-Za-z]+$/)) {
return 0;
}
if (n == 1)
{
return str[0].charCodeAt() - '0' .charCodeAt();
}
return (10 * myAtoiRecursive(str, n - 1) + str[n - 1].charCodeAt() - '0' .charCodeAt());
}
let str = "112" ;
let n = str.length;
document.write(myAtoiRecursive(str, n));
</script>
|
Time complexity: O(n),
Auxiliary Space: O(n)
Last Updated :
19 Sep, 2023
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