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.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.

spoilers alert… click to show requirements for atoi.

Requirements for atoi:
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

C++ STL算法系列2—find ，find_first_of ， find_if ， adjacent_find的使用

分析：可以使用string的成员函数转换，但是很多条件没考虑，会导致不能通过，故自己实现了其转换功能；

int main()
{
    string s;
    while (cin >> s)
    {
        long long tmp;   //作为一个临时存储，并于后面的比较
        int n;
        vector<int>ivec; //存放从字符串中取出的数
        if (s.empty())tmp = 0; //如果字符串为空，直接返回0
        else
        {
            /*此处是用到C++的字符串的成员函数s.find_first_of()，返回位置,<泛型库中也有find_first_of函数，返回迭代器>但是很多情况无法考虑到，不好实现，所以没采用，但是一般字符串到值的转换是可以实现的*/
            /*auto pos = s.find_first_of("+-0123456789"); //取得第一次出现"+-0123456789"中任何字符的第一个位置
            if (pos == string::npos)tmp = 0; //如果在s中没有找到"+-0123456789"中任何一个字符，则返回string::npos
            else
            {
                tmp = stoll(s.substr(pos)); //找到之后，则进行转换，直到遇到不是数字字符的数终止，比如“+345a89”，转换为345
                if (tmp > 2147483647) tmp = 0; //超出范围
                if (tmp*(-1)>2147483648) tmp = 0;
            }*/ 
            int i=0;
            int flag=0;
            int count = 0;
            while (s[i] == ' ')i++; //去除字符串开始的空格（“   0111”），其实cin可以自动去除前面的空格，但是leetcode上面给的字符串需要自己去除空格
            while (i < s.size()) 
            { 


                if (count > 1)return 0;  //count的作用是排除这种情况例如：+-2，这种情况的话，直接是返回0，当遇到+或-时，count++，则第二次遇到的话，count>1;
                if (s[i] == '+'){ flag = 0; count++; } //+,flag标记为0
                else
                    if (s[i] == '-'){ flag = 1; count++; }
                    else  //如果出现+或-的话，后面跟的是数字，如果再跟其他字符的话，则要终止while，比如：+100013a89，则后面的89不会存入容器
                        if (s[i] >= '0'&&s[i] <= '9')
                        {
                            ivec.push_back(s[i] - '0');
                        }
                        else break;
                i++;


            }
            int pos = ivec.size();
            long long degree = 1;
            tmp = 0;
            while (--pos >= 0)//从后面向前面一次从容器中取出数字，然后乘以degree，degree变化（1，10，100，1000.....）
            {
                tmp += ivec[pos] * degree;
                if (flag == 0 && tmp>2147483647) //如果tmp>int的最大正数2147483647，则直接取2147483647
                {
                    tmp = 2147483647;
                    break;
                }
                else
                    if (flag == 1 && tmp > 2147483648)//如果tmp<int表示的最小负数-2147483648时，则直接取-2147483648
                    {
                        tmp = 2147483648;
                        break;
                    }
                degree *= 10;
            }
            if (flag == 0)n = tmp; //最后将tmp给到n；
            else if (flag == 1)n = tmp*(-1);
        }
        cout << n << endl;
    }
}