Implementatoito 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 theC++function had been updated. If you still see your function signature accepts aconst char *argument, please click the reload buttonto 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.
按照题目的要求来就是如下4点：
1.数字前面有空格，如s=“ 123”，需要把空格去掉。
2.数字前出现正负号，，要考虑符号，如s=“ +123” , s=“ -123”。
3.数字中出现了不必要的字符，返回字符前的数字，也即数字字符不属于[0,9]，如s=“ 123a123”。
4.数字越界，超过了范围（-2147483648–2147483647)，若超过了负数的，输出-2147483648，超过了正数的输出2147483647。
class Solution {public:int myAtoi(string str) {if(str.length() == 0)return 0;//索引记录int i = 0;//剔除空格for(i = 0; i < str.length();){if(str[i] == ' ')i++;elsebreak;}//记录正负符号int sign = 1;if(str[i] == '-'){sign = -1;i++;}else if(str[i] == '+')i++;//转成数字long long ans = 0;while (str[i] >= '0' && str[i] <= '9'){ans = ans*10 + (str[i]-'0');//越界判断if(ans > INT_MAX)return sign < 0 ? INT_MIN : INT_MAX;i++;}ans *= sign;return (int)ans;}};
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