8. String to Integer (atoi)

8. String to Integer (atoi)

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Analysis:
这道题确实不太好做，很难想到那么全面。
参考：http://blog.csdn.net/ljiabin/article/details/40508889

学了个新函数：

cout<<std::numeric_limits<int>::max()<<endl;  int最大值

cout<<std::numeric_limits<int>::min()<<endl;   int最小值

Source Code（C++）:

#include <iostream>
#include <vector>
#include <string>
using namespace std;

class Solution {
public:
    int myAtoi(string str) {        
        if (str.empty()){
            return 0;
        }
        //去除前边的空格
        while(str.at(0) == ' '){ 
            str.erase(0, 1);
        }
        //判断正负号并取出前边的有效数字
        string::size_type k;
        int start_pos = 0;
        int num_sign=0;
        if (str.at(0)=='+' || str.at(0)=='-') {
            k=str.find_first_not_of("0123456789", 1);
            start_pos = 1;
            num_sign = str.at(0)=='+' ? 1:-1;
        }
        else {
            k=str.find_first_not_of("0123456789");
            start_pos = 0;
            num_sign = 1;
        }
        if (k==string::npos){
            k=str.size();
        }
        //转换为数字
        long long num=0;
        for (int i=start_pos; i<k; i++){
            num = num*10 + num_sign*(int)(str.at(i)-'0');
            if (num>(int)2147483647 || num<(int)-2147483648){  //此处不加(int)会出错，应该是计算机内部数据存储方式的问题，可尝试/*cout << (long long)-2147483648;*/
                return num>0 ? 2147483647:-2147483648;
            }
        }
        return num;
    }
};

int main() {
    Solution sol;
    cout << sol.myAtoi("+1");
    cout << sol.myAtoi("10.10");    
    cout << sol.myAtoi("010");  
    cout << sol.myAtoi("+-0");  
    cout << sol.myAtoi(" 1a33");
    cout << sol.myAtoi("-21474836486");
    return 0;
}