[leetcode]34 String to Integer (atoi)

题目链接：https://leetcode.com/problems/string-to-integer-atoi/
Runtimes：15ms

1、问题

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.

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.

2、分析

这是一道很烦人的题目，题目不难，边界处理超级变态。将字符串转换成为数字，几点要求：1、有符号时，符号后面一连串数字就是需要转换的数字，如 ” -1010023630o4”，数字为“-1010023630”；2、无符号时，以数字非0开始的一连串数字为需要转换的数字” 0 0 0134”，数字转换为“134”；3、如果出现多个符号，这返回0，如“+-1928387”，数字返回0；4、超过 “-2147483648”，返回 “-2147483648”，超过 “2147483647”，返回 “2147483647”。

3、总结

认真思考，不要气馁。

4、实现

class Solution {public:    int check(int a, int b, char f)    {        if (f == '-')        {            if ((INT_MIN + b) / 10 == (0 - a)){                if (b >= 8)                    return INT_MIN;            }else if ((INT_MIN + b) / 10 > (0 - a))                return INT_MIN;        }        else{            if ((INT_MAX - b) / 10 <= a)                return INT_MAX;        }        return 0;    }    int atoi(string str) {        int x = 0;        char flag = ' ';        if ("" == str)            return x;        int i = 0;        while (str[i] < '0' || str[i] > '9')        {            if (str[i] == '-' || str[i] == '+')            {                if (flag == ' ')                    flag = str[i];                else                    return 0;            }            else            if (flag == '-' || flag == '+')                return 0;            else            if (str[i] != ' ')                return 0;            i++;        }        for (; i < str.length(); i++)        {            if (str[i] < '0' || str[i] > '9')                break;            int temp = check(x, (str[i] - '0'), flag);            if (temp != 0)                return temp;            x = (x * 10 + (str[i] - '0'));        }        if (flag == '-')            x = 0 - x;        return x;    }};

5、反思

学会了溢出的边界处理，最终还是得数学来解决。