String to Integer (atoi)

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.

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.

Tips:

即将String转换成Integer。用trim去掉头尾的空格后根据正负号更新sign，res设为-1。需要注意的是，进入循环时，每一步更新res后都要判断是否大于或小于Integer_MAX，如果超出范围及时返回。如果这一位不是数字也需要及时返回。

Code：

public class Solution {
    public int myAtoi(String str) {
        if (str == null || str.length() == 0) return 0;
        str = str.trim();
        int sign = 1;
        long res = 0;
        int index = 0;
        StringBuilder sb = new StringBuilder();
        if (str.charAt(0) == '+') index++;
        else if (str.charAt(0) == '-') {
            sign = -1;
            index++;
        }
        while (index < str.length() && Character.isDigit(str.charAt(index))) {
            res = res * 10 + (str.charAt(index++) - '0');
            if (res * sign > Integer.MAX_VALUE) return Integer.MAX_VALUE;
            if (res * sign < Integer.MIN_VALUE) return Integer.MIN_VALUE;
        }
        res = res * sign;
        return (int)res;
    }
}