请你来实现一个 myAtoi(string s) 函数，使其能将字符串转换成一个 32 位有符号整数（类似 C/C++ 中的 atoi 函数）。
函数 myAtoi(string s) 的算法如下：
0 。必要时更改符号（从步骤 2 开始）。[−231, 231 − 1] ，需要截断这个整数，使其保持在这个范围内。具体来说，小于 −231 的整数应该被固定为 −231 ，大于 231 − 1 的整数应该被固定为 231 − 1 。注意：
\n\n' ' 。\n\n
示例 1：
\n\n\n输入：s = \"42\"\n输出：42\n解释：加粗的字符串为已经读入的字符，插入符号是当前读取的字符。\n第 1 步：\"42\"（当前没有读入字符，因为没有前导空格）\n ^\n第 2 步：\"42\"（当前没有读入字符，因为这里不存在 '-' 或者 '+'）\n ^\n第 3 步：\"42\"（读入 \"42\"）\n ^\n解析得到整数 42 。\n由于 \"42\" 在范围 [-231, 231 - 1] 内，最终结果为 42 。\n\n
示例 2：
\n\n\n输入：s = \" -42\"\n输出：-42\n解释：\n第 1 步：\" -42\"（读入前导空格，但忽视掉）\n ^\n第 2 步：\" -42\"（读入 '-' 字符，所以结果应该是负数）\n ^\n第 3 步：\" -42\"（读入 \"42\"）\n ^\n解析得到整数 -42 。\n由于 \"-42\" 在范围 [-231, 231 - 1] 内，最终结果为 -42 。\n\n\n
示例 3：
\n\n\n输入：s = \"4193 with words\"\n输出：4193\n解释：\n第 1 步：\"4193 with words\"（当前没有读入字符，因为没有前导空格）\n ^\n第 2 步：\"4193 with words\"（当前没有读入字符，因为这里不存在 '-' 或者 '+'）\n ^\n第 3 步：\"4193 with words\"（读入 \"4193\"；由于下一个字符不是一个数字，所以读入停止）\n ^\n解析得到整数 4193 。\n由于 \"4193\" 在范围 [-231, 231 - 1] 内，最终结果为 4193 。\n\n\n
\n\n
提示：
\n\n0 <= s.length <= 200s 由英文字母（大写和小写）、数字（0-9）、' '、'+'、'-' 和 '.' 组成\n\n
注意：本题与主站 8 题相同：https://leetcode-cn.com/problems/string-to-integer-atoi/
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