leetcode 8 String to Integer (atoi)

题目：

https://leetcode.com/problems/string-to-integer-atoi/description/

题意：

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.

实现atoi函数

思路：

丢弃前面的空白字符，从第一个非空字符开始，有可能是′+′或者′−′，然后取后面的数字，直到结尾或者遇到非数字字符。如果得到的数字是负数且小于INT_MIN，那么就返回INT_MIN，若是正数且大于INT_MAX，那么返回INT_MAX，否则就返回本来的值

代码：

class Solution {public:    int myAtoi(string str) {        string::size_type id = 0;        while(str[id] == ' ' || str[id] == '\t')            ++id;        int f = 1;        if(str[id] == '-' || str[id] == '+')        {            if(str[id] == '-')                f = -f;            ++id;        }        long long ans = 0;        for(int i = id; str[i]; ++i)        {            if(! isdigit(str[i]))                break;            ans = ans * 10 + str[i]-'0';            if(ans*f < INT_MIN)                return INT_MIN;            if(ans*f > INT_MAX)                return INT_MAX;        }        return ans * f;    }};