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.

思路：题目不难，但是很多情况要考虑清楚，"",　　"+",　　"     +0 123",　　"123ab2",以及越界的情况

class Solution {
public:
    string getInteger(string &str)
    {
        int idx=0;
        for(idx=0; idx<str.size(); ++idx)
        {
            if(str[idx]<'0' || str[idx]>'9')break;
        }
        return str.substr(0, idx);
    }
    int myAtoi(string str) {
        if(str.size()==0)return 0;
        int idx=0;
        while(idx<str.size() && str[idx]==' ')++idx;//开头先去空格
        str=str.substr(idx, str.size()-idx);
        bool isActive=true;
        if(str[0]>'9' || str[0]<'0')
        {
            if(str[0]!='+' && str[0]!='-')return 0;
            if(str[0]=='-')isActive=false;
            str=str.substr(1, str.size()-1);//去正负号
        }
        string validStr=getInteger(str);//得到有效数字
        if(validStr.size()==0)return 0;
        if(validStr.size()>10)//int型，长度大于10肯定越界
        {
            if(isActive)return INT_MAX;
            else return INT_MIN;
        }
        long n=0;
        for(int idx=0; idx<validStr.size(); ++idx)
        {
            n=n*10+validStr[idx]-'0';
        }
        if(!isActive)n=-n;
        if(n>INT_MAX)return INT_MAX;
        if(n<INT_MIN)return INT_MIN;
        return n;
    }
};