LeetCode_String to Integer

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String to Integer (atoi)

Implement atoi which converts a string to an integer.
（字符串转32位整形）

Note:

Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: \([−2^{31}, 2^{31} − 1]\). For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Example:



字符串过滤-正则表达式

首先删除字符串首尾的空格，通过正则表达式过滤剩下以 “+-0123456789” 开头的句子，提取剩下的字符串首部的数字串，再将其转换为整形。（注意需要将超出范围的部分返回0。）

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import re

class Solution:
    def myAtoi(self, str):
        """
        :type str: str
        :rtype: int
        """
        string = str.strip()
        if not re.match(r'^(\-|\+)?\d+', string):
            return 0
       
        INT_MIN = -pow(2, 31)
        INT_MAX = pow(2, 31) - 1
        char_string = '+-0123456789'
        
        num_string = ''
        for i, char in enumerate(string):
            if char not in char_string:
                break
            if i != 0 and (char == '+' or char == '-'):
                break
            num_string += char
       
        if num_string[0] == '-':
            result = int(num_string[1:])
            result = max(result * - 1, INT_MIN)
            return result
        elif num_string[0] == '+': 
            result = int(num_string[1:])
        else:
            result = int(num_string)
        result = min(result, INT_MAX)  
        return result

注：需要考虑 \(+/- \) 出现在数字字符串中间的部分，因此 ‘0-1’ 需要通过if i != 0 and (char == '+' or char == '-')过滤掉。当然，也可以直接通过match的group()函数，其返回值为匹配的字符串，进而获取开始的数值字符串。

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num_string = re.match(r'^(\-|\+)?\d+', string).group()