Leetcode 8 - 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.

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.

细节实现题，主要注意以下几点：
1 - 跳过字符串首的空格；
2 - 判断正负
3 - 考虑整型溢出问题

class Solution {public:    int myAtoi(string str) {        int result = 0;        int sign = 1;        int i=0;        //跳过字符串首的空格        while(str[i]==' ') i++;        //判断正负        if(str[i]=='-'){            sign = -1;            i++;        }else if(str[i]=='+'){            i++;        }        for(;i<str.length();i++){            if(str[i]<'0' || str[i]>'9') return result*sign;            //溢出条件            if(result>INT_MAX/10 || (result==INT_MAX/10 && str[i]-'0'>INT_MAX%10)) return sign==-1?INT_MIN:INT_MAX;            result = result*10 + str[i]-'0';        }        return result * sign;    }};