(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;
     }
 };