String to Integer (atoi)

描述
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.
分析
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”

代码

public class StringtoInteger {
    static int INT_MAX = 2147483647;
    static int INT_MIN = -2147483648;

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        String str = "-39x24x8fc";
        System.out.println(atoi(str));
    }
    // LeetCode, String to Integer (atoi)
    // 时间复杂度 O(n)，空间复杂度 O(1)

    public static int atoi(String str) {
        int num = 0;
        int sign = 1;
        int n = str.length();
        int i = 0;
        while (str.charAt(i) == ' ' && i < n)
            i++; // 去空格
        if (str.charAt(i) == '+')
            i++; // 判符号
        if (str.charAt(i) == '-') { // 判符号-
            sign = -1; // sign符号
            i++;
        }
        for (; i < n; i++) { //
            if (str.charAt(i) < '0' || str.charAt(i) > '9')
                // break; //跳出本循环 得到部分数字
                continue; // 跳出本次循环，不运行循环内的下面程序，然后跳到下次循环。得到全部数字
            if (num > INT_MAX / 10 || 
                (num == INT_MAX / 10 && (str.charAt(i) - '0') > INT_MAX % 10)) { // 判超限
                return sign == -1 ? INT_MIN : INT_MAX;
            }
            num = num * 10 + str.charAt(i) - '0';
        }
        return num * sign;
    }

}