Leetcode : String to Integer (atoi)

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本文介绍了一个简单的atoi函数实现方法，用于将字符串转换为整数，并详细讨论了输入处理、正负号判断、越界判断等问题。

String to Integer (atoi)Dec 27 '11

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.

Note：该题也比较简单，但是也要注意有几点需要注意。1) 正负号的判断；2) 越界判断；3) 空格符的判断；4) 数字前几位为0，如00123，但是我发现测试用例中并没有包含这种情况(可能我想多了)

代码如下：

class Solution {
public:
    int atoi(const char *str)
    {
        const char* pChar = str;
        
        double dRet = 0;
        int bFlag = 1;
        while (pChar)
        {
            if (*pChar == ' ')
            {
                ++pChar;
                continue;
            }
            
            if (*pChar == '-')
            {
                bFlag = -1;
                ++pChar;
            }
            else if (*pChar == '+')
                ++pChar;
            
            break;
        }
        
        while (pChar)
        {   
            if (*pChar < '0' || *pChar > '9')
            {
                break;
            }
            
            if (dRet != 0)
                dRet = dRet * 10 + *pChar - '0';
            else
                dRet = *pChar - '0';
                
            ++pChar;
        }
        
        dRet *= bFlag;
        
        if (dRet > INT_MAX)
            dRet = INT_MAX;
        else if (dRet < INT_MIN)
            dRet = INT_MIN;
            
        return dRet;
        
    }
};