8.Atoi

题目：

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.

class Solution {
public:
    int myAtoi(string str) {
        
    }
};

题目大意：给定一个字符串，让它转换为数字；对字符串的输入情况做以下几种考虑：

1、字串为空或者全是空格，返回0； 

2、忽略掉前缀空格后，遇到的第一个字符，如果是‘+’或‘－’号或者是数字，则开始处理数字,否则返回0；

3、处理数字的过程中，如果之后的字符非数字，就停止转换，返回当前值；

4、 在上述处理过程中，如果转换出的值超出了范围INT_MAX(2147483647)和 INT_MIN(-2147483648)则返回INT_MAX， INT_MIN。

class Solution {
public:
	int atoi(const char *str) {
		if (str == NULL)   
			return 0; 
		int i = 0;
		while (str[i] != '\0' && str[i] == ' ')
			i++;
		if (str[i] == '\0')
			return 0;
		int flag = 1;
		if (str[i] == '+')
		    i++;
		else if (str[i] == '-')
		{
			flag = -1;
			i++;
		} 
		long long result = 0;
		while (str[i] != '\0')
		{
			if (str[i] >= '0' && str[i] <= '9')
				result = result * 10 + flag * (str[i] - '0');
			else
				return result;
			if (result  >  INT_MAX || result <  INT_MIN)   //溢出处理  
				return result  > 0 ? INT_MAX : INT_MIN;
			i++;
		}
		return result;
	}
};

用到stl中的find_first_not_of()函数后，就省去了上述的一些步骤，代码如下：

int myAtoi(string str)
{
	long result = 0;
	int indicator = 1;
	for (int i = 0; i < str.size();)
	{
		i = str.find_first_not_of(' ');
		if (str[i] == '-' || str[i] == '+')
			indicator = (str[i++] == '-') ? -1 : 1;
		while ('0' <= str[i] && str[i] <= '9')
		{
			result = result * 10 + (str[i++] - '0');
			if (result*indicator >= INT_MAX) return INT_MAX;
			if (result*indicator <= INT_MIN) return INT_MIN;
		}
		return result*indicator;
	}
}