String to Integer (atoi)，integeratoi

String to Integer (atoi)，integeratoi

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.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button  to reset your code definition.

spoilers alert... click to show requirements for atoi.

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.

問題：將字符竄轉換成數字
分析：感覺題目不難，但是細節很多，容易想不到
1.數字前面有空格 如s=“    123456”
2.數字前出現了不必要或多於的字符導致數字認證錯誤，輸出0   如s=“   b1234”  ，s=“  ++1233” , s=“ +-1121”
3.數字中出現了不必要的字符，返回字符前的數字 如s=“   12a12” ， s=“ 123  123”
4.數字越界 超過了范圍（-2147483648--2147483647) 若超過了負數的 輸出-2147483648  超過了正數的輸出2147483647
在科普一個知識點，倘若某個數超過了2147483647則會變為負數，反過來一樣  

 1 int myAtoi(char* str) {
 2     int sign = 1;           //設置符號位,初始設置1為正數 如果前面沒有符號就代表正數
 3     long long sum = 0;             //sum范圍要大於int
 4     if(str == NULL)
 5         return 0;
 6     while(*str == ' ') str++;         //去掉前面空格
 7     if(*str == '-'){                   
 8         sign = -1;
 9         str++;
10     }                                     //取符號
11     else if(*str == '+'){
12         sign = 1;
13         str++;
14     }
15     while(*str != '\0'){
16         if(*str < '0' || *str > '9')              //不符合規則的取消，如“  +-123”
17             break;
18         else if(*str >= '0' && *str <= '9'){
19             sum = sum * 10 + *str -48;
20             if(sum > 2147483648)                       //sum范圍超出
21                 break;
22         }
23         str++;
24     } 
25     if(sign == 1 && sum > 2147483647)
26         return 2147483647;
27     if(sign == -1 && sum > 2147483648)
28         return -2147483648;
29     sum = (int)sum * sign;                                //最後把sum轉化成int類型
30     return sum;
31 }