Leetcode - Strobogrammatic Number

A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

Write a function to determine if a number is strobogrammatic. The number is represented as a string.

For example, the numbers "69", "88", and "818" are all strobogrammatic.
[分析]
把数字颠倒180度后看仍然一样的有0,1,8三个，另外69,96这种颠倒后也一样，需特殊处理。
因此从中间往两端或者从两端往中间逐个判断对称下标处字符，注意长度为奇数时要处理中间元素。讨论区有牛人分享了4行代码的方案，awesome！
[ref]
https://leetcode.com/discuss/50594/4-lines-in-java

public class Solution {
    // Method 2
    public boolean isStrobogrammatic(String num) {
        if (num == null) return false;
        for (int i = 0, j = num.length() - 1; i <= j; i++, j--) {
            if (!"00 11 88 696".contains(num.charAt(i) + "" + num.charAt(j)))
                return false;
        }
        return true;
    }
    // Method 1
    private HashSet<Character> upDownSameDigit = new HashSet<Character>();
    public Solution() {
        upDownSameDigit.add('0');
        upDownSameDigit.add('1');
        upDownSameDigit.add('8');
    }
    public boolean isStrobogrammatic1(String num) {
        if (num == null || num.length() == 0)
            return false;
        int mid = num.length() / 2;
        if (num.length() % 2 == 1) {
            if (!upDownSameDigit.contains(num.charAt(mid)))
                return false;
            return checkSymmetrically(num, mid - 1, mid + 1);
        } else {
            return checkSymmetrically(num, mid - 1, mid);
        }
    }
    public boolean checkSymmetrically(String num, int i, int j) {
        while (i >= 0) {
            char left = num.charAt(i);
            char right = num.charAt(j);
            boolean okCase1 = left == right && upDownSameDigit.contains(left);
            boolean okCase2 = (left == '6' && right == '9') || (left == '9' && right == '6');
            if (!okCase1 && !okCase2)
                return false;
            i--; j++;
        }
        return true;
    }
}