8. String to Integer (atoi)

Problem：

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.

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.

题意挺明确的，就是要自己去实现一个atoi函数，然后注意各种不合法或者不规则的样例。例如题目提示有可能会以很多的空格开头，所以可以先在一开始清除空格。还得判断开头的正号和负号。一旦在转换数字的时候遇到非数字，后面的都可以不用管了。所以其实就是遍历一边字符串，然后判断当前字符的情况，时间复杂度为O(n)， 空间复杂度为O(1)。

Code:(LeetCode运行19ms)

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