(letcode)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.

问题：将字符窜转换成数字
分析：感觉题目不难，但是细节很多，容易想不到
1.数字前面有空格如s=“ 123456”
2.数字前出现了不必要或多于1个的字符导致数字认证错误，输出0 如s=“ b1234” ，s=“ ++1233” , s=“ +-1121”
3.数字中出现了不必要的字符，返回字符前的数字如s=“ 12a12” ， s=“ 123 123”
4.数字越界超过了范围（-2147483648--2147483647) 若超过了负数的输出-2147483648 超过了正数的输出2147483647
在科普一个知识点，倘若某个数超过了2147483647则会变为负数，反过来一样

 class Solution {

 public:

     int myAtoi(string str) {

         long long result = ;

         int i = ;

         int flag1 = ;

         int flag2 = ;

         if(str.empty()) return ;

         while(str[i] != '\0' && str[i] == ' ') i++;

         while(str[i] == '-') { flag1++;i++; }

         while(str[i] == '+') { flag2++;i++; }

         while(i < str.length())

         {

             if(str[i] >= '' && str[i] <= '')

             {

                 if(flag1 > ||flag2> || (flag1 == &&flag2 == ))

                 {

                     result = ;

                     break;

                 }

                 else if(flag1 == )

                 {

                     result = result * - (str[i]-'');

                     if(result < INT_MIN) result = INT_MIN;

                 }

                 else

                 {

                     result = result * + (str[i]-'');

                     if(result > INT_MAX) result = INT_MAX;

                 }

                 i++;

             }

             else

             {

                 break;

             }

         }

         int num = (int)result;

         return num;

     }

 };

巴特西

(letcode)String to Integer (atoi)

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