leetCode3.3(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.

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.

这是一道很经典的面试题，思路并不复杂，但是考虑的东西比较多。

细节题。注意几个测试用例：

1. 不规则输入，但是有效，”-3924x8fc”，” + 413”,

2. 无效格式，” ++c”, ” ++1”

3. 溢出数据，”2147483648”

public static int solution_3_3(String str){int num=0;int sign=1;final int max_int=2147483647;int i=0;while(str.charAt(i)==' '&&i<str.length())i++;if(str.charAt(i)=='+')i++;if(str.charAt(i)=='-'){i++;sign=-1;}while(i<str.length()){if(str.charAt(i)<'0'||str.charAt(i)>'9')break;if((sign==1)&&((num>max_int/10)||(num==max_int/10&&(str.charAt(i)-'0')>max_int%10))){return max_int;}if((sign==-1)&&((num>max_int/10)||(num==max_int/10&&(str.charAt(i)-'0')>(max_int%10+1)))){return -(max_int+1);}num=num*10+(str.charAt(i)-'0');i++;}return sign*num;}