leetcode 刷题之路 44 String to Integer (atoi)

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.

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.

将一个字符串转化为整数。

要求：

1.无效字符串或者空字符串或者全为空白字符串返回0。

2.如果字符串数字字符前有除‘+’,'-'和空白字符的其他字符，算作是无效字符串，返回0。

3.数字字符后面出现的其他字符忽略掉

5.如果数字超过整型变量的边界值，返回边界值。

面试常考题目，明确了题目要求，程序不难实现。

AC CODE：

class Solution {public:    int atoi(const  char *str)     {        if(str==NULL)                   return 0;        int sign=1,i=0;        while(str[i]==' ')            i++;        if(str[i]=='+'||str[i]=='-')            sign=str[i++]=='+'?1:-1;        else if(!isdigit(str[i]))            return 0;        long long sum=0;        while(isdigit(str[i]))        {            sum=str[i++]-'0'+sum*10;            if(sign==1&&sum>=INT_MAX)                return INT_MAX;            if(sign==-1&&sum-1>=INT_MAX)                return INT_MIN;        }        return sign*sum;    }};