翻译

实现“atoi”将字符串转换成整型数。

提示：仔细考虑所有可能的输入。如你想要挑战，请不要参阅下面并问问自己都有哪些可能的输入请看。

说明：模糊的指定（没有给定的输入规格）就是为了这个问题。你负责收集所有可能的输入。

atoi的要求：

函数首先放弃尽可能多的空字符直到找到一个非空白字符。然后从这个字符开始，带上可选的初始加/减字符，其后还可能跟着越多越好的数字，并将它们解释成一个数值。

这个字符串可能在这些数字之后包含一些附加的字符，它们可以可以被忽略，并对函数的行为没有影响。

如果字符串str中第一个非空格的序列不是一个有效的整型数，或者因为str为空或仅有空格字符而不存在这样一个序列，那么不执行任何转换。

如果可以不执行任何有效的转换，则返回零值。如果正确的值在值域范围之外，则返回INT_MAX（2147483647）或INT_MIN（-2147483647）。

原文

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Requirements for atoi:
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

英语渣渣实在没看懂题目，不知道有哪些条件，于是就慢慢写代码，根据报错继续改……结果代码改到了80行……还是不能完成所有条件，还是从网上荡了一个下来，来日再战！

（好吧，在写博客，也就上面的翻译过程中，我发现题目懂了……）

public class Solution
{
    public int MyAtoi(string str)
    {
        if (string.IsNullOrEmpty(str))
        {
            return 0;
        }
        var result = 0;
        var i = 0;
        // clean all the whitespaces in the beginning
        while (i < str.Length && str[i] == ' ')
        {
            i++;
        }
        // check positive or negative sign
        var sign = 1;
        switch (str[i])
        {
            case '-':
                sign = -1;
                i++;
                break;
            case '+':
                sign = 1;
                i++;
                break;
        }
        // check the rest of numbers
        while (i < str.Length && str[i] >= '0' && str[i] <= '9')
        {
            // check overflow
            try
            {
                checked
                {
                    result = result * 10 + (str[i++] - '0');
                }
            }
            catch (OverflowException)
            {
                return sign == 1 ? int.MaxValue : int.MinValue;
            }
        }
        return sign * result;
    }
}

1045 / 1045 test cases passed.
Status: Accepted
Runtime: 168 ms
Your runtime beats 16.85% of csharp submissions.