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

//Author: yqtao
//Email : [email protected]
//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 2
int 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;
}
//test
int main()
{
    string s = " 2147483648 ";
    cout << myAtoi(s)<< endl;
}