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.

唉·，没想到这么简单一个都这么难写。

注意四点情况：

１.输入为空

２.正负号

３.数据溢出

４.空白格

数据溢出不好弄，下面这个方法是利用INT_MAX/10的值和当前result未乘10的值进行比较，判断是否溢出。由于溢出正负极值分别为2147483647   -2147483648，所以为了简练，有两种情况：(1)直接大于INT_MAX/10，说明足够大了，绝对溢出了（2）等于INT_MAX/10，可能会溢出，我们需要再加一个条件判断，是否最后一位大于7。对于正数，2147473648即为最小的溢出值，最后一位大于7。对于负值，如果值为2147483648，虽然没溢出，但是刚好，假设它的值到了这个档次，即时溢出了我们也要返回-2147483648，所以这种情况也应该返回。

class Solution {public:    int myAtoi(string str) {        if(str.size() == 0)            return 0;                    int result = 0;        int sign = 1;        int i = 0;                    while(str[i] == ' ')            ++i;                    if(str[i] == '-' || str[i] == '+')            sign = (str[i++] == '-') ? -1 : 1;        while(str[i] >= '0' && str[i] <= '9'){            if(result > INT_MAX / 10 || (result == INT_MAX / 10 && str[i]-'0' > 7)){                return sign == 1 ? INT_MAX : INT_MIN;            }            result = result * 10 + str[i++] - '0';        }           return result * sign;    }   };