leetcode8. String to Integer (atoi)

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本文介绍了一个将字符串转换为整数（atoi）的C++实现方法，详细解析了如何处理字符串中的空白字符、正负号及数字字符，并讨论了溢出情况的处理。

medium程度题

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.

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.

跳过空格，确定正负，每次计算前都要判断当前的sum值乘以10后是否会溢出

AC解：

class Solution {
public:
    int myAtoi(string str) 
    {
        int num = 0;
        int sign = 1;
        const int len = str.length();
        int i = 0;
        while (str[i] == ' ' && i < len)//跳过空格
            i++;
        if (str[i] == '+')
            i++;
        else if (str[i] == '-')		//确定正负
        {
            sign = -1;
            i++;
        }
        
        while (i < len)
        {
            if (str[i] < '0' || str[i] > '9')
                break;
            if (num > INT_MAX / 10 || (num == INT_MAX / 10 && //判断下次计算是否会溢出
                                            (str[i] - '0') > INT_MAX % 10))
                return sign == -1 ? INT_MIN : INT_MAX;
            num = num * 10 + str[i++] - '0';
        }
        
        return num * sign;
    }
};