## [8\. String to Integer (atoi)](https://leetcode.com/problems/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 is 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:**

**Input:** s = "42"

**Output:** 42

**Explanation:**

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:**

**Input:** s = " -042"

**Output:** -42

**Explanation:**

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:**

**Input:** s = "1337c0d3"

**Output:** 1337

**Explanation:**

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)
^

**Example 4:**

**Input:** s = "0-1"

**Output:** 0

**Explanation:**

Step 1: "0-1" (no characters read because there is no leading whitespace)
^
Step 2: "0-1" (no characters read because there is neither a '-' nor '+')
^
Step 3: "0-1" ("0" is read in; reading stops because the next character is a non-digit)
^

**Example 5:**

**Input:** s = "words and 987"

**Output:** 0

**Explanation:**

Reading stops at the first non-digit character 'w'.

**Constraints:**

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

class Solution {
public:
    int myAtoi(string s) {
        // 紀錄正負號，預設是正號
        int sign = 1;
        int res = 0; 
        int i = 0;
        int n = s.size();
        // 掃 s，略過空白格
        while (i < n && s[i] == ' ') i++;

        // 紀錄正負號，i++
        if (i < n && s[i] == '+') {
            sign = 1;
            i++;
        } else if (i < n && s[i] == '-') {
            sign = -1;
            i++;
        }
        // 現在的 i 就是整數的起始位置
        while (i < n && isdigit(s[i])) {
            int digit = s[i] - '0';
            // 檢查有沒有超過 32-bit signed integer 的範圍
            if ((res > INT_MAX / 10) || (res == INT_MAX / 10 && digit > INT_MAX % 10)) { 
                return sign == 1 ? INT_MAX : INT_MIN;
            }
            // 10 進位
            res = 10 * res + digit;
            i++;
        }
        return sign * res;
    }
};

:::success
- 時間複雜度：$O(N)$
- 空間複雜度：$O(1)$
:::