[LeetCode Javascript]8 - MyAtoI

Problem

https://leetcode.com/problems/string-to-integer-atoi/description/

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 the next 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.

Solution

My Original Solution

First, I just wanted to understand the question, so I wrote straightforward code.

Full Code

const myAtoi = function(s) {
    s = s.trim();
    let multiplier = 1;
    let numberString = '';
    if (s[0] === '-'){
        multiplier = -1;
        s = s.substring(1);
    } else if (s[0] === '+'){
        s = s.substring(1);
    }
    
    for(let i = 0; i < s.length; i++){
        const c = s[i];
        if(!isNaN(c*1) && c !==" "){
            numberString += c;
        } else{
            break;
        }
    }
    if(numberString === ''){
        return 0;
    }
    let output = numberString * multiplier;
    if(output > 2**31 - 1){
        return 2**31 - 1;
    }
    if(output < -(2**31)){
        return -(2**31)
    }
    return output
};

Explain

s = s.trim();

In the first line, I trimmed whitespaces to deal with leading whitespaces.

if (s[0] === '-'){
    multiplier = -1;
    s = s.substring(1);
} else if (s[0] === '+'){
    s = s.substring(1);
}

I use else if because there are the cases like ’-+83’, in which the return value should be 0.

if(output > 2**31 - 1){
    return 2**31 - 1;
}
if(output < -(2**31)){
    return -(2**31)
}

The return value should be a 32-bit signed integer. So I added the above code, but I wish it was cleaner.

Result and Time Complexity.

Time complexity is $O(n)$. Because there is only one loop.