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.

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.

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

 

int myAtoi(char* str) {
    int sign = 1;           //设置符号位,初始设置1为正数 如果前面没有符号就代表正数
    long long sum = 0;             //sum范围要大于int
    if(str == NULL)
        return 0;
    while(*str == ' ') str++;         //去掉前面空格
    if(*str == '-'){                   
        sign = -1;
        str++;
    }                                     //取符号
    else if(*str == '+'){
        sign = 1;
        str++;
    }
    while(*str != ''){
        if(*str < '0' || *str > '9')              //不符合规则的取消，如“  +-123”
            break;
        else if(*str >= '0' && *str <= '9'){
            sum = sum * 10 + *str -48;
            if(sum > 2147483648)                       //sum范围超出
                break;
        }
        str++;
    } 
    if(sign == 1 && sum > 2147483647)
        return 2147483647;
    if(sign == -1 && sum > 2147483648)
        return -2147483648;
    sum = (int)sum * sign;                                //最后把sum转化成int类型
    return sum;
}