LeetCode 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.

分析：
考虑各种边界情况。
” 000123” return 123
” +000123” return 123
“-123” return -123
“123a4x” return 123

越界情况：
“2147483647” return 2147483647
“-2147483648” return -2147483648
“2147483648” return 2147483647
“-2147483649” return 2147483647
越界情况可以比较中间结果是否已经达到2147483640，即INT_MAX/10。然后再判断下一位值的情况，进而判断是否越界。

class Solution {public:    int myAtoi(string str)     {        int result = 0;        int i = 0;        int sign = 1;        if (str[i] != '\0')        {            while (str[i] == ' ') ++i;            if (str[i] == '+') ++i;            else if (str[i] =='-') sign = -1, ++i;            while (str[i] == '0') ++i;            if (str[i] != '0')            {                while (str[i] != '\0')                {                    if (str[i] >= '0' && str[i] <= '9')                    {                        if (result > INT_MAX / 10) return sign == -1 ? INT_MIN : INT_MAX;  // 越界情况                        if (result == INT_MAX / 10) {                            if ((str[i] - '0' <= 7 && sign == 1) || (str[i] - '0' <= 8 && sign == -1))                            {                                result = (result * 10) + (str[i++] - '0');                                return sign * result;                            }                            return sign == -1 ? INT_MIN : INT_MAX;  // 越界情况                        }                        result = (result * 10) + (str[i++] - '0');                    }                    else {                        break;                    }                }            }        }        return sign * result;    }};