LeetCode: String to Integer (atoi)

Implement atoi which converts a string to an integer.

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.

Note:

• Only the space character ’ ’ is considered as whitespace character.
• Assume we are dealing with an environment which could only store
  integers within the 32-bit signed integer range: [−2³¹, 2³¹ − 1]. If
  the numerical value is out of the range of representable values,
  INT_MAX (2³¹ − 1) or INT_MIN (−2³¹) is returned.

Example 1:

Input: “42”
Output: 42

Example 2:

Input: " -42"
Output: -42
Explanation: The first non-whitespace character is ‘-’, which is the minus sign.
Then take as many numerical digits as possible, which gets 42.

Example 3:

Input: “4193 with words”
Output: 4193
Explanation: Conversion stops at digit ‘3’ as the next character is not a numerical digit.

Example 4:

Input: “words and 987”
Output: 0
Explanation: The first non-whitespace character is ‘w’, which is not a numerical
digit or a +/- sign. Therefore no valid conversion could be performed.

Example 5:

Input: “-91283472332”
Output: -2147483648
Explanation: The number “-91283472332” is out of the range of a 32-bit signed integer.
Thefore INT_MIN (−231) is returned.

把字符串转化为整数，规则比较多，先去除空格，接下去第一个字符可以是正负号，下一个字符如果不是数字就返回0，是数字的话就取出最长的连续数字型字符串。

1. 用strip()方法去除空格
2. 判断是否为空字符串或者‘-’和‘+’，是的话就返回0
3. 判断第一个字符是否为正负号
4. 遍历后续字符串，是数字就添加到新字符串里，遇到非数字便break
5. 新字符串转换为int
6. 判断是否越界
7. 返回结果

class Solution(object):
    def myAtoi(self, str):
        """
        :type str: str
        :rtype: int
        """
        str = str.strip()

        if str == '' or str == '-' or str == '+':
            return 0

        flag = 1
        res = ''

        if str[0] == '-':
            flag = -1
            str = str[1:]
        elif str[0] == '+':
            str = str[1:]

        if not str[0].isdigit():
            return 0

        for c in str:
            if c.isdigit():
                res += c
            else:
                break

        res = int(res) * flag

        if res > 2147483647:
            return 2147483647
        if res < -2147483648:
            return -2147483648

        return res

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