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.

该题最好先考虑各种输入情况，再动手写代码

输入：

空串

125

12.36

546a

#4655

     45656

-0001

2147483647

-2147483648

9999999999999999

-99999999999999

 1 class Solution { 2 public: 3     int atoi(const char *str) { 4         if( !str ) return 0; 5         while( *str && *str == ‘ ‘ ) ++str; //寻找第一个非空格字符 6         bool minus = false; 7         if( *str == ‘-‘ ) { //判断是否为负 8             minus = true; 9             ++str;10         }11         else if( *str == ‘+‘ ) ++str;   //或‘+‘12         long long ans = 0;  //为防止溢出，使用long long13         while( *str && isdigit(*str) ) {    //碰到非数字或结束符就结束14             ans = ans * 10 + (*str - ‘0‘);15             ++str;16         }17         ans = minus ? -ans : ans;   //还原真实值18         if( ans > INT_MAX ) return INT_MAX; //处理上溢19         else if( ans < INT_MIN ) return INT_MIN;    //处理下溢20         else return ans;21     }22 };