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

又是一道細(xì)節(jié)多過(guò)算法的題。有好多坑爹的地方需要考慮啊。
常規(guī)的：

• 開頭和結(jié)尾的space需要忽略
• 正負(fù)號(hào)只能出現(xiàn)在開頭一次

非常規(guī)的

• 遇到空格或者非法字符，拋棄后面的所有字符，而不是返回INVALID。
• 對(duì)于INVALID情況，返回0。
• 對(duì)于溢出，正數(shù)返回INT_MAX，負(fù)數(shù)返回INT_MIN。

用int sign還蠻巧妙的，比用boolean表示正負(fù)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);
    }
}