LeetCode第八题（String to Integer (atoi)）

String to Integer (atoi)

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

• 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.

Python代码：

class Solution:
    def myAtoi(self, str):
        """
        :type str: str
        :rtype: int
        """
        if (str == ""):
            return 0
        MAX = 0x7FFFFFFF
        MIN = 0x80000000
        length = len(str)
        ret = 0
        op = 1
        p = 0
        while (str[p] == ' ' and (p + 1) < length):
            p += 1
        if (str[p] == '+' or str[p] == '-'):
            if (str[p] == '-'):
                op = -1
            p += 1
        for i in range(p, length):
            if ('0' <= str[i] and str[i] <= '9'):
                ret = ret * 10 + int(str[i])
                if (op == 1 and ret > MAX):
                    return MAX
                if (op == -1 and ret > MIN):
                    return -MIN
            else:
                break
        return int(ret * op)

总结：

• 空字符串返回0
• 全空格字符串返回0，此处需要注意边界
• 处理字符串前先去除空格
• 判断有无符号，最多只能允许一个符号
• 对每个字符乘以位的权值再加起来，就转换为相应的数值，此时要判断是否超出int的边界（0x7FFFFFFF 与 0x80000000），过界了就返回边界值
• 当字符串中出现非数字字符时即数字字符读取结束，退出循环（123 1，123x1都返回123）