LeetCode 8. String to Integer (atoi)简单易懂的解法

//Author: yqtao//Email : [e-mail]//Date  :2016.7.5/*************************************** Implement atoi to convert a string to an integer.* 实现一个字符串到数字的转换******************************//*   this problems is difficult at should check  INT_MAX (2147483647) or INT_MIN (-2147483648)    it can be  complex very we have no idea;   for this have two solutions to do:*/#include<iostream>#include<string>#include<cctype>using namespace std;//solution 1:int myAtoi(string str){    int len = str.size(), i = 0;    if (len < 1) return 0;    while (isspace(str[i])) i++;  //skip the space    int indicator = 1;      if (str[i] == '+' || str[i] == '-') {        indicator = (str[i++] == '-') ? -1 : 1;    }    long long  result = 0;       //using long long is very simple to solve     while ('0' <= str[i] && str[i] <= '9')   //can also use isdigit(s[i])    {        int digit = str[i] - '0';        result = result * 10 + digit;        if (result*indicator >= INT_MAX) return INT_MAX;   //you see it is very easy        if (result*indicator <= INT_MIN) return INT_MIN;        i++;    }    return result*indicator;  }//solution 2int myAtoi1(string str){    int len = str.size(), i = 0;    if (len < 1)        return 0;    while (isspace(str[i])) i++;  //skip the space    bool neg = false;    if (str[i] == '-' || str[i] == '+') {        neg = (str[i] == '-');        i++;    }    int  result = 0;    while (isdigit(str[i]))    {        int digit = str[i] - '0';        if (neg) {            if (-result < (INT_MIN + digit) / 10) {    //it can be difficult to understand                return INT_MIN;            }        }        else {            if (result >(INT_MAX - digit) / 10) {                return INT_MAX;            }        }        result = result * 10 + digit;        i++;    }    return  neg ? -result : result;}//testint main(){    string s = "  2147483648   ";    cout << myAtoi(s)<< endl;}