题解 | #把字符串转换成整数(atoi)#

class Solution {
public:
 
    const int min = pow(2,31) * (-1);
    const int max = pow(2,31) - 1;

    int StrToInt(string s) {
        int sign = 0;   //符号
        int len = s.length();   //字符串长度
        int signcount = 0;   //符号个数
        vector<int> rec;
        for(int i =0;i < len;i++){
            if(s[i] == '+' || s[i] == '-'){
                signcount ++;
                if(s[i] == '-') sign = 1;    //符号置1
                if(signcount > 1) return 0;    //符号个数大于1个，出错返回0
            }
            else if(s[i] == '0' || s[i] == '1' || s[i] == '2' || s[i] == '3' || s[i] == '4' || s[i] == '5' || s[i] == '6' || s[i] == '7' || s[i] == '8' || s[i] == '9'){
                rec.push_back(s[i] - '0');  //前面只有空格和符号，碰到数字插入数字即可
                if(i+1 <len && s[i+1] != '0' && s[i+1] != '1' && s[i+1] != '2' && s[i+1] != '3' && s[i+1] != '4' && s[i+1] != '5' && s[i+1] != '6' && s[i+1] != '7' && s[i+1] != '8' && s[i+1] != '9') break;   //数字的下一位不是数字，说明当前合法数字已提取完毕，可以退出
            }
            else if(s[i] == ' ') continue;   //如果碰到数字前是空格，就继续
            else break;    //碰到数字前，遇到非符号，非空格  结束
        }
        if(rec.size() == 0) return 0;
        long sum = 0;
        for(auto x : rec){
            sum = sum * 10 + x;
            if(sum > max){    //超过最大值最小值，返回最大值或最小值
                if(sign == 1) return min;
                else return max;
            }
        }
        if(sign == 1) return sum * (-1);
        else return sum;
    }
};