[C++]LeetCode 8:String to Integer (atoi)（字符串转int）

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.

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.

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.

分析：

重写atoi是面试宝典中经常出现的一道题了。主要的问题有以下几个：

1、空格的排除

2、非数字字符的判断

3、溢出问题

其中溢出问题已在中介绍过[C++]LeetCode 7:Reverse Integer（翻转整数），与此处使用的溢出情况一致，故不再介绍。

AC Code(C++)：

class Solution {    public:    //1045 / 1045 test cases passed.    //Runtime: 18 ms        #define INT_MAX 2147483647    #define INT_MIN (-INT_MAX - 1)        int myAtoi(string str) {    if (str.empty())return 0;//字符串为空情况    int begin = 0;    while (begin < str.size() && str[begin] == ' ')++begin;//跳过空格    if (begin == str.size())return 0;         int flag = 1;//设置正负指示值    int num = 0;    if (str[begin] == '+'){    ++begin;    }    else if (str[begin] == '-'){    flag = -1;    ++begin;    }        while (begin < str.size()){    if (str[begin] < '0' || str[begin] > '9')return num;//非数字字符处理    if ((num != 0 && (INT_MAX / abs(num) < 10)) || ((unsigned int)abs(num * 10) + (unsigned int)(str[begin] - '0') > INT_MAX)){//溢出    if (flag == 1){    return INT_MAX;    }    else{    return INT_MIN;    }    }    num = num * 10 + flag * (str[begin] - '0');    ++begin;    }    return num;    }};