Leetcode - Strobogrammatic Numbe

My code:

public class Solution {
    private int counter = 0;
    public int strobogrammaticInRange(String low, String high) {
        char[][] pairs = new char[][]{{'0', '0'}, {'1', '1'}, {'6', '9'}, {'8', '8'}, {'9', '6'}};
        for (int i = low.length(); i <= high.length(); i++) {
            helper(0, i - 1, new char[i], pairs, low, high);
        }
        return counter;
    }
    
    private void helper(int begin, int end, char[] board, char[][] pairs, String low, String high) {
        if (begin > end) {
            String s = String.valueOf(board);
            if (s.length() == low.length() && s.compareTo(low) < 0) {
                return;
            }
            else if (s.length() == high.length() && s.compareTo(high) > 0) {
                return;
            }
            else {
                counter++;
            }
        }
        else if (begin == end) {
            board[begin] = '0';
            helper(begin + 1, end - 1, board, pairs, low, high);
            board[begin] = '1';
            helper(begin + 1, end - 1, board, pairs, low, high);
            board[begin] = '8';
            helper(begin + 1, end - 1, board, pairs, low, high);
        }
        else {
            for (int i = 0; i < pairs.length; i++) {
                char x = pairs[i][0];
                char y = pairs[i][1];
                if (begin == 0 && x == '0') {
                    continue;
                }
                board[begin] = x;
                board[end] = y;
                helper(begin + 1, end - 1, board, pairs, low, high);
            }
        }
    }
    
    public static void main(String[] args) {
        Solution test = new Solution();
        int ret = test.strobogrammaticInRange("50", "100");
        System.out.println(ret);
    }
}

reference:
https://discuss.leetcode.com/topic/31386/easiest-20ms-94-java-solution

一开始我想的是，能不能直接穷举出来。
但是发现不太容易，于是看答案。发现所看的答案，也是拿 II 作为基础，把 长度 [low.length(), high.length()] 的可能找出来，然后删去那些超出范围的。
当然，整个过程可以更加优化。使得空间复杂度一直保持在 O(high.length())

Anyway, Good luck, Richardo! -- 09/22/2016