String to Integer (atoi) LeetCode Solution

Problem – String to Integer (atoi) LeetCode Solution

Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++’s atoi function).

The algorithm for myAtoi(string s) is as follows:

1. Read in and ignore any leading whitespace.
2. Check if the next character (if not already at the end of the string) is '-' or '+'. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
3. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.
4. Convert these digits into an integer (i.e. "123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).
5. If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -231 should be clamped to -231, and integers greater than 231 - 1 should be clamped to 231 - 1.
6. Return the integer as the final result.

Note:

• Only the space character ' ' is considered a whitespace character.
• Do not ignore any characters other than the leading whitespace or the rest of the string after the digits.

Example 1:

Input: s = "42"
Output: 42
Explanation: The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42" (no characters read because there is no leading whitespace)
         ^
Step 2: "42" (no characters read because there is neither a '-' nor '+')
         ^
Step 3: "42" ("42" is read in)
           ^
The parsed integer is 42.
Since 42 is in the range [-231, 231 - 1], the final result is 42.

Example 2:

Input: s = "   -42"
Output: -42
Explanation:
Step 1: "   -42" (leading whitespace is read and ignored)
            ^
Step 2: "   -42" ('-' is read, so the result should be negative)
             ^
Step 3: "   -42" ("42" is read in)
               ^
The parsed integer is -42.
Since -42 is in the range [-231, 231 - 1], the final result is -42.

Example 3:

Input: s = "4193 with words"
Output: 4193
Explanation:
Step 1: "4193 with words" (no characters read because there is no leading whitespace)
         ^
Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
         ^
Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
             ^
The parsed integer is 4193.
Since 4193 is in the range [-231, 231 - 1], the final result is 4193.

Constraints:

• 0 <= s.length <= 200
• s consists of English letters (lower-case and upper-case), digits (0-9), ' ', '+', '-', and '.'.

String to Integer (atoi) LeetCode Solution in Java

public int myAtoi(String str) {
    int index = 0, sign = 1, total = 0;
    //1. Empty string
    if(str.length() == 0) return 0;

    //2. Remove Spaces
    while(str.charAt(index) == ' ' && index < str.length())
        index ++;

    //3. Handle signs
    if(str.charAt(index) == '+' || str.charAt(index) == '-'){
        sign = str.charAt(index) == '+' ? 1 : -1;
        index ++;
    }
    
    //4. Convert number and avoid overflow
    while(index < str.length()){
        int digit = str.charAt(index) - '0';
        if(digit < 0 || digit > 9) break;

        //check if total will be overflow after 10 times and add digit
        if(Integer.MAX_VALUE/10 < total || Integer.MAX_VALUE/10 == total && Integer.MAX_VALUE %10 < digit)
            return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;

        total = 10 * total + digit;
        index ++;
    }
    return total * sign;
}

String to Integer (atoi) LeetCode Solution in Python

class Solution(object):
    def myAtoi(self, s):
        """
        :type str: str
        :rtype: int
        """
        ###better to do strip before sanity check (although 8ms slower):
        #ls = list(s.strip())
        #if len(ls) == 0 : return 0
        if len(s) == 0 : return 0
        ls = list(s.strip())
        
        sign = -1 if ls[0] == '-' else 1
        if ls[0] in ['-','+'] : del ls[0]
        ret, i = 0, 0
        while i < len(ls) and ls[i].isdigit() :
            ret = ret*10 + ord(ls[i]) - ord('0')
            i += 1
        return max(-2**31, min(sign * ret,2**31-1))

String to Integer (atoi) LeetCode Solution in Python3

class Solution:
    def myAtoi(self, str: str) -> int:
        value, state, pos, sign = 0, 0, 0, 1

        if len(str) == 0:
            return 0

        while pos < len(str):
            current_char = str[pos]
            if state == 0:
                if current_char == " ":
                    state = 0
                elif current_char == "+" or current_char == "-":
                    state = 1
                    sign = 1 if current_char == "+" else -1
                elif current_char.isdigit():
                    state = 2
                    value = value * 10 + int(current_char)
                else:
                    return 0
            elif state == 1:
                if current_char.isdigit():
                    state = 2
                    value = value * 10 + int(current_char)
                else:
                    return 0
            elif state == 2:
                if current_char.isdigit():
                    state = 2
                    value = value * 10 + int(current_char)
                else:
                    break
            else:
                return 0
            pos += 1

        value = sign * value
        value = min(value, 2 ** 31 - 1)
        value = max(-(2 ** 31), value)

        return value

String to Integer (atoi) LeetCode Solution in C++

int myAtoi(string str) {
    long result = 0;
    int indicator = 1;
    for(int i = 0; i<str.size();)
    {
        i = str.find_first_not_of(' ');
        if(str[i] == '-' || str[i] == '+')
            indicator = (str[i++] == '-')? -1 : 1;
        while('0'<= str[i] && str[i] <= '9') 
        {
            result = result*10 + (str[i++]-'0');
            if(result*indicator >= INT_MAX) return INT_MAX;
            if(result*indicator <= INT_MIN) return INT_MIN;                
        }
        return result*indicator;
    }
}

String to Integer (atoi) LeetCode Solution in Swift

class Solution {
    func myAtoi(_ s: String) -> Int {
        guard !s.contains("+ ") else { return 0 }
        let val = (s as NSString).integerValue
        return val >= Int32.max ? Int(Int32.max) : max(Int(Int32.min), val)
    }
}

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