Leetcode NO.8 String to Integer (atoi)

本题题目要求如下：

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.

本题如果不看requirement直接写出来还是有一定难度的，但是有了提示，就不算太难，基本思路就是如果碰到空格，自动往后移，然后检查正负号，之后就是数字，数字记得每次乘之前看看是不是在本次运算会溢出。。这题的思路很简单，难点是细节处理，并且想到一种将细节处理简单化的方法。。我开始写的代码很复杂，后来看论坛，根据排名第一的答案，重写了一遍，现在就比较简洁了，代码如下：

class Solution {public:    int myAtoi(string str) {        int sign = 1;       // positive: +1; negative: -1;        int i = 0;        int base = 0;        while (str[i] == ' ') {            ++i;        }        if (str[i] == '+' or str[i] == '-') {            sign = (str[i++] == '+' ? 1 : -1);        }        while (i < str.length() and isdigit(str[i])) {            if (base > INT_MAX / 10 or (base == INT_MAX / 10 and str[i] > '7')) {                return (sign == 1 ? INT_MAX : INT_MIN);            }            base = 10 * base + str[i] - '0';            ++i;        }        return sign * base;    }};