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class Solution {
public:
    int myAtoi(string s) {
        int res = 0, sign = 1, i = 0, n = s.size();
        while(s[i] == ' ') i ++;                          //处理空格
        if(s[i] == '-'){                                  //处理符号
            sign = -1;
            i ++;
        }
        else if(s[i] == '+') i ++;

        while(i < n && isdigit(s[i])){                    //处理数字 
            if(res > INT_MAX / 10 || res == INT_MAX / 10 && (s[i] - '0' > 7)){ //即将溢出
                return sign == 1 ? INT_MAX : INT_MIN;
            }
            res = res * 10 + (s[i] - '0');
            i++; 
        }
        return res * sign;
    }
};