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.

题意

实现atoi

1 数字开头可能有空格

2 有符号

3 中间有字母处理字母之前的数字

4 有溢出 输出最大/最小

5 比较大小的话用 longlong 就没问题了

代码:

class Solution:
    # @return an integer
    def atoi(self, str):
        str = str.strip()
        newStr = []
        for i in xrange(len(str)):
            if '0' <= str[i] <= '9' or (str[i] in ('+', '-') and i == 0):
                newStr.append(str[i])
            else:
                break
        if newStr in ([], ['+'], ['-']):
            return 0
        elif -2147483648 <= int(''.join(newStr)) <= 2147483647:
            return int(''.join(newStr))
        elif int(''.join(newStr)) > 2147483647:
            return 2147483647
        else:
            return -2147483648