Medium
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:
'-' 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."123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).[-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.Note:
' ' is considered a whitespace character.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.
Example 4:
Input: s = “words and 987”
Output: 0
Explanation:
Step 1: "words and 987" (no characters read because there is no leading whitespace)
         ^
Step 2: "words and 987" (no characters read because there is neither a '-' nor '+')
         ^
Step 3: "words and 987" (reading stops immediately because there is a non-digit 'w')
         ^
The parsed integer is 0 because no digits were read.
Since 0 is in the range [-231, 231 - 1], the final result is 0.
Example 5:
Input: s = “-91283472332”
Output: -2147483648
Explanation:
Step 1: "-91283472332" (no characters read because there is no leading whitespace)
        ^
Step 2: "-91283472332" ('-' is read, so the result should be negative)
          ^
Step 3: "-91283472332" ("91283472332" is read in)
                     ^
The parsed integer is -91283472332.
Since -91283472332 is less than the lower bound of the range [-231, 231 - 1], the final result is clamped to -231 = -2147483648.
Constraints:
0 <= s.length <= 200s consists of English letters (lower-case and upper-case), digits (0-9), ' ', '+', '-', and '.'.class Solution {
    fun myAtoi(str: String): Int {
        if (str.isEmpty()) {
            return 0
        }
        var i = 0
        var negativeSign = false
        val input = str.toCharArray()
        while (i < input.size && input[i] == ' ') {
            i++
        }
        if (i == input.size) {
            return 0
        } else if (input[i] == '+') {
            i++
        } else if (input[i] == '-') {
            i++
            negativeSign = true
        }
        var num = 0
        while (i < input.size && input[i] <= '9' && input[i] >= '0') {
            // current char
            var tem = input[i] - '0'
            tem = if (negativeSign) -tem else tem
            // avoid invalid number like 038
            if (num == 0 && tem == '0'.code) {
                i++
            } else if (num == Int.MIN_VALUE / 10 && tem <= -8 || num < Int.MIN_VALUE / 10) {
                return Int.MIN_VALUE
            } else if (num == Int.MAX_VALUE / 10 && tem >= 7 || num > Int.MAX_VALUE / 10) {
                return Int.MAX_VALUE
            } else {
                num = num * 10 + tem
                i++
            }
        }
        return num
    }
}