The atoi() function in C takes a string (which represents an integer) as an argument and returns its value of type int. So basically the function is used to convert a string argument to an integer.
Syntax:
int atoi(const char strn)
Parameters: The function accepts one parameter strn which refers to the string argument that is needed to be converted into its integer equivalent.
Return Value: If strn is a valid input, then the function returns the equivalent integer number for the passed string number. If no valid conversion takes place, then the function returns zero.
Now let’s understand various ways in which one can create their own atoi() function supported by various conditions:
Approach 1: Following is a simple implementation of conversion without considering any special case.
- Initialize the result as 0.
- Start from the first character and update result for every character.
- For every character update the answer as result = result * 10 + (s[i] – ‘0’)
Java
// A simple Java program for// implementation of atoiclass GFG { // A simple atoi() function static int myAtoi(String str) { // If str is NULL or str contains non-numeric // characters then return 0 as the number is not // valid if (str == "" || str.matches("[a-zA-Z]+") || str.matches(".*[0-9].*")) { return 0; } // Initialize result int res = 0; // Iterate through all characters // of input string and update result // take ASCII character of corresponding digit and // subtract the code from '0' to get numerical // value and multiply res by 10 to shuffle // digits left to update running total for (int i = 0; i < str.length(); ++i) res = res * 10 + str.charAt(i) - '0'; // return result. return res; } // Driver code public static void main(String[] args) { String str = "8f9789"; // Function call int val = myAtoi(str); System.out.println(val); }}// This code is contributed by PrinciRaj1992 |
89789
Time Complexity : O(N), as only one traversal of the string is needed.
Auxiliary Space : O(1) , as no extra space is required.
Approach 2: This implementation handles the negative numbers. If the first character is ‘-‘ then store the sign as negative and then convert the rest of the string to number using the previous approach while multiplying sign with it.
Java
// Java program for// implementation of atoiclass GFG { // A simple atoi() function static int myAtoi(char[] str) { // Initialize result int res = 0; // Initialize sign as positive int sign = 1; // Initialize index of first digit int i = 0; // If number is negative, then // update sign if (str[0] == '-') { sign = -1; // Also update index of first // digit i++; } // Iterate through all digits // and update the result for (; i < str.length; ++i) res = res * 10 + str[i] - '0'; // Return result with sign return sign * res; } // Driver code public static void main(String[] args) { char[] str = "-123".toCharArray(); // Function call int val = myAtoi(str); System.out.println(val); }}// This code is contributed by 29AjayKumar |
-123
Time Complexity: O(n), where n is the length of the input string ‘str’. The algorithm iterates through all the characters of the string once to convert them to integer digits.
Space Complexity: O(1). The algorithm uses constant extra space to store the integer result, sign, and the index of the first digit. Therefore, the space complexity is O(1).
Approach 3: This implementation handles various type of errors. If str is NULL or str contains non-numeric characters then return 0 as the number is not valid.
-134
Approach 4: Four corner cases needs to be handled:
- Discards all leading whitespaces
- Sign of the number
- Overflow
- Invalid input
To remove the leading whitespaces run a loop until a character of the digit is reached. If the number is greater than or equal to INT_MAX/10. Then return INT_MAX if the sign is positive and return INT_MIN if the sign is negative. The other cases are handled in previous approaches.
Dry Run:
Below is the implementation of the above approach:
Java
// A simple Java program for// implementation of atoiclass GFG { static int myAtoi(char[] str) { int sign = 1, base = 0, i = 0; // if whitespaces then ignore. while (str[i] == ' ') { i++; } // sign of number if (str[i] == '-' || str[i] == '+') { sign = 1 - 2 * (str[i++] == '-' ? 1 : 0); } // checking for valid input while (i < str.length && str[i] >= '0' && str[i] <= '9') { // handling overflow test case if (base > Integer.MAX_VALUE / 10 || (base == Integer.MAX_VALUE / 10 && str[i] - '0' > 7)) { if (sign == 1) return Integer.MAX_VALUE; else return Integer.MIN_VALUE; } base = 10 * base + (str[i++] - '0'); } return base * sign; } // Driver code public static void main(String[] args) { char str[] = " -123".toCharArray(); // Function call int val = myAtoi(str); System.out.printf("%d ", val); }}// This code is contributed by 29AjayKumar |
-123
Complexity Analysis for all the above Approaches:
- Time Complexity: O(n).
Only one traversal of string is needed. - Space Complexity: O(1).
As no extra space is required.
Exercise:
Write your won atof() that takes a string (which represents an floating point value) as an argument and returns its value as double.
Please refer complete article on Write your own atoi() for more details!
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