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 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，将字符串转换成整数。
atoi方法首先将任何前置的空格舍弃，直到发现第一个不是空格的字符。然后从这个字符开始，第一个可能是正负号，也可能没有，后面紧跟数字。最后把这个字符串解析成数字。 （要考虑正负号）
字符串可能在整数数字后有一些其他的字符，我们需要忽略这些字符，并且这个不影响函数的功能。
如果一开始的非空格字符串不是有效的完整的数字，或者因为str是空或者只包含空格，则转换不会发生。
如果没有有效的转换发生，函数会返回0。如果正确的值超出能表示的数字的范围，则返回最大整数（2147483647）或者最小整数（-2147483648）

/*
1. 字串为空或者全是空格，返回0；
2. 字串的前缀空格需要忽略掉；
3. 忽略掉前缀空格后，遇到的第一个字符，如果是‘+’或‘－’号，继续往后读；如果是数字，则开始处理数字；如果不是前面的2种，返回0；
4. 处理数字的过程中，如果之后的字符非数字，就停止转换，返回当前值；
5. 在上述处理过程中，如果转换出的值超出了int型的范围，就返回int的最大值或最小值。

*/

class Solution {public:    int myAtoi(string str) {        // IMPORTANT: Please reset any member data you declared, as          // the same Solution instance will be reused for each test case.          if (str.size()==0)   //空字串             return 0;          //忽略前缀空格          int i = 0;          while (str[i] != '\0' && str[i] == ' ')            ++i;          if (str[i] == '\0')            return 0;          int max = INT_MAX; //表示最大数         int min = INT_MIN;  //表示最小数        int signal = 1;          //处理+、-号          if (str[i] == '+')          {              signal = 1;              ++i;          }          else if (str[i] == '-')          {              signal = -1;              ++i;          }          //转换整数          long long sum = 0;          while (str[i] != '\0')          {              if (str[i] >= '0' && str[i] <= '9')                   sum = sum * 10 + signal * (str[i] - '0');  //每次得到的数都用signal表示出来。            else                   return sum;              if (sum > max || sum < min)   //溢出处理                 return sum > 0 ? max : min;              ++i;          }          return sum;      }};