String to Integer (atoi)--LeetCode

String to Integer (atoi)

（原题链接：点击打开链接）

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. 

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.

Solution：

找到第一个非空白字符并且判断是否为有效的整数字符（‘+’ ‘-’ ‘0~9’），若无效，则返回0。若第一个非空白字符为正负号，则记录数值的正负，然后从第二个字符开始记录值，直至找到无效的整数字符或者到达字符串尾部，若超出INT的取值范围，则返回INT_MAX或INT_MIN。最后若原数值为负数，则将数值变为负。

注意循环内的判断溢出，防止数目太大造成result的溢出而造成答案的错误。

class Solution {public:    int myAtoi(string str) {        long long int result = 0;        int i = 0;        int lessthan0 = 0;        for(i = 0; i < str.length(); i++)        {            if (str[i] != ' ') break;        }        string substr = str.substr(i);        if(substr.length() == 0 || (substr[0] != '-' && substr[0] != '+' && (substr[0] < '0' ||substr[0] > '9'))) return 0;        if(substr[0] == '-')        {            lessthan0 = 1;            substr = substr.substr(1);        }        else if(substr[0] == '+')        {            substr = substr.substr(1);        }        for(i = 0; i < substr.length(); i++)        {            if(substr[i] < '0' || substr[i] > '9') break;            result = result * 10 + (substr[i] - '0');            if(result > INT_MAX) break;        }        if(result > INT_MAX)        {            return lessthan0 ? INT_MIN : INT_MAX;        }        return lessthan0 ? -1 * result : result;    }};