LeetCode-String to Integer(atoi)

1. String to Integer (atoi)（Medium）

Description
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.

Analysis
题目大意就是把字符串转换为数字，只不过题目的要求与标准的atoi函数并不完全一样，例如不合法的输入返回0而不是-1。另外就是需要考虑到各种情况。可以用多种测试样例测试，就可以清楚题目的要求。

代码：

class Solution {public:    int myAtoi(string str) {        if (str.empty())   //空字符串            return 0;        //忽略前缀空格        int i = 0;        while (str[i] != '\0' && str[i] == ' ')            ++i;        if (str[i] == '\0')            return 0;        int signal = 1;        //处理+、-号        if (str[i] == '+') {            signal = 1;            ++i;        } else if (str[i] == '-') {            signal = -1;            ++i;        }        long long sum = 0;        while (str[i] != '\0') {            if (str[i] >= '0' && str[i] <= '9')                sum = sum * 10 + signal * (str[i] - '0');            else                return (int)sum;            if (sum > INT_MAX || sum < INT_MIN)   //溢出处理                return sum > 0 ? INT_MAX : INT_MIN;            ++i;        }        return (int)sum;    }};