Leetcode: String to Integer (atoi)

class Solution {
public:
    int myAtoi(string str) {
        if (str == "") return 0; //判断是否为空
        int i = 0, sign = 1;
        long long sum = 0; //用long long来存结果，易于判断是否溢出，用int的话比较麻烦
        while (str[i] == ' ') i++; //忽略前置的空格
        //判断是否溢出，溢出的话根据sign判断返回最小值还是最大值，这里体现了用longlong的好处
        if (str[i] == '+' || str[i] == '-') sign = 44 - str[i++];
        for (i; i < str.length(); i++) {
            if (str[i] >= '0' && str[i] <= '9') sum = sum * 10 + str[i] - '0';
            else break;
            //判断是否溢出，溢出的话根据sign判断返回最小值还是最大值，这里体现了用longlong的好处
            if (sum > INT_MAX) return sign == -1 ? INT_MIN : INT_MAX;
        }
        return sum * sign;
    }
};