8. String to Integer (atoi)

Description

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 *
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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.

Solution

Iterative

又是一道细节多过算法的题。有好多坑爹的地方需要考虑啊。
常规的：

• 开头和结尾的space需要忽略
• 正负号只能出现在开头一次

非常规的

• 遇到空格或者非法字符，抛弃后面的所有字符，而不是返回INVALID。
• 对于INVALID情况，返回0。
• 对于溢出，正数返回INT_MAX，负数返回INT_MIN。

用int sign还蛮巧妙的，比用boolean表示正负好。

class Solution {
    public int myAtoi(String str) {
        if (str == null || str.isEmpty()) {
            return 0;
        }
        int n = str.length();
        int i = 0;

        // escape leading spaces
        if (str.charAt(i) == ' ') {
            while (i < n && str.charAt(i) == ' ') {
                ++i;
            }
        }
        if (i == n) {
            return 0;
        }

        int sign = 1;
        if (str.charAt(i) == '+' || str.charAt(i) == '-') {
            sign = (str.charAt(i++) == '+') ? 1 : -1;
        }

        long res = 0;   // use long to handle overflow
        while (i < n && str.charAt(i) >= '0' && str.charAt(i) <= '9') {
            res = res * 10 + (str.charAt(i++) - '0');
            if (res * sign >= Integer.MAX_VALUE) return Integer.MAX_VALUE;
            if (res * sign <= Integer.MIN_VALUE) return Integer.MIN_VALUE;
        }

        return (int) (res * sign);
    }
}