8. String to Integer (atoi)

Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer.

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

1. Whitespace: Ignore any leading whitespace (" ").
2. Signedness: Determine the sign by checking if the next character is '-' or '+', assuming positivity if neither present.
3. Conversion: Read the integer by skipping leading zeros until a non-digit character is encountered or the end of the string is reached. If no digits were read, then the result is 0.
4. Rounding: If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then round the integer to remain in the range. Specifically, integers less than -231 should be rounded to -231, and integers greater than 231 - 1 should be rounded to 231 - 1.

Return the integer as the final result.

Example 1:

The underlined characters are what is read in and 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)
           ^

Example 2:

Step 1: "   -042" (leading whitespace is read and ignored)
            ^
Step 2: "   -042" ('-' is read, so the result should be negative)
             ^
Step 3: "   -042" ("042" is read in, leading zeros ignored in the result)
               ^

Example 3:

Step 1: "1337c0d3" (no characters read because there is no leading whitespace)
         ^
Step 2: "1337c0d3" (no characters read because there is neither a '-' nor '+')
         ^
Step 3: "1337c0d3" ("1337" is read in; reading stops because the next character is a non-digit)
             ^

Constraints:

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

Approach 01: Iterative Parsing

class Solution {
public:
    int myAtoi(string s) {
        int i = 0, n = s.length();
        while (i < n && s[i] == ' ') i++;
        int sign = 1;
        if (i < n && (s[i] == '+' || s[i] == '-')) {
            sign = (s[i] == '-') ? -1 : 1;
            i++;
        }
        long res = 0;
        while (i < n && isdigit(s[i])) {
            res = res * 10 + (s[i] - '0');
            if (res * sign >= INT_MAX) return INT_MAX;
            if (res * sign <= INT_MIN) return INT_MIN;
            i++;
        }
        return (int)(res * sign);
    }
};

class Solution:
    def myAtoi(self, s: str) -> int:
        s = s.strip()
        if not s: return 0
        sign = 1
        i = 0
        if s[i] == '+':
            i += 1
        elif s[i] == '-':
            sign = -1
            i += 1
        res = 0
        while i < len(s) and s[i].isdigit():
            res = res * 10 + int(s[i])
            i += 1
        res *= sign
        INT_MAX = 2**31 - 1
        INT_MIN = -2**31
        if res > INT_MAX: return INT_MAX
        if res < INT_MIN: return INT_MIN
        return res

class Solution {
    public int myAtoi(String s) {
        int i = 0, n = s.length();
        while (i < n && s.charAt(i) == ' ') i++;
        if (i == n) return 0;
        int sign = 1;
        if (s.charAt(i) == '+' || s.charAt(i) == '-') {
            sign = (s.charAt(i) == '-') ? -1 : 1;
            i++;
        }
        long res = 0;
        while (i < n && Character.isDigit(s.charAt(i))) {
            res = res * 10 + (s.charAt(i) - '0');
            if (res * sign >= Integer.MAX_VALUE) return Integer.MAX_VALUE;
            if (res * sign <= Integer.MIN_VALUE) return Integer.MIN_VALUE;
            i++;
        }
        return (int)(res * sign);
    }
}