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.

分析： 非常重要且出名的题目. 有非常多的边界情况要考虑。根据上面的spoiler,已经基本可以将整个流程写出来，但是注意INT_MAX 和 INT_MIN不同，这就造成了当符号不同时，边界不同需要注意

给出代码：

 1 class Solution { 2 public: 3 int atoi(const char *str) { 4     int num = 0; 5     bool sign = false; 6     const int n = strlen(str); 7     int i = 0; 8     while (str[i] == ‘ ‘ && i < n) i++; 9     if (str[i] == ‘+‘) i++;10     else if (str[i] == ‘-‘) {11         sign = true;12         i++;13     }14     for (; i < n; i++) {15         int rec = str[i]-‘0‘;16         if (str[i] < ‘0‘ || str[i] > ‘9‘)17             break;18         if (num > INT_MAX / 10)19         {20             return sign?INT_MIN:INT_MAX;21         }22         23         if(num == INT_MAX / 10)24         {25             if(sign && rec>(INT_MAX%10+1)) return INT_MIN;26             else if(!sign && rec>INT_MAX%10) return INT_MAX;27         }28         num = num * 10 + str[i] - ‘0‘;29     }30     if(sign) num = - num;31     return num;32 }33 };