易混淆数 II

/*
 * 题目思路：
 * 通过Java Stream API，我们可以使用递归生成所有可能的数字，
 * 并使用流的过滤功能来筛选出所有有效的易混淆数。
 */
import java.util.*;
import java.util.stream.*;

public class ConfusingNumberIIStream {
    private static final char[] DIGITS = {'0', '1', '6', '8', '9'};
    private static final Map<Character, Character> CONFUSING_MAP = Map.of(
        '0', '0',
        '1', '1',
        '6', '9',
        '8', '8',
        '9', '6'
    );

    public long confusingNumberII(int N) {
        return generateNumbers(0, new StringBuilder())
            .filter(num -> num > 0 && num <= N && isConfusingNumber(Long.toString(num)))
            .count();
    }

    private Stream<Long> generateNumbers(long cur, StringBuilder sb) {
        if (cur > Integer.MAX_VALUE) return Stream.empty();

        Stream<Long> result = Stream.of(cur);
        for (char digit : DIGITS) {
            if (cur == 0 && digit == '0') continue;
            sb.append(digit);
            result = Stream.concat(result, generateNumbers(cur * 10 + (digit - '0'), sb));
            sb.deleteCharAt(sb.length() - 1);
        }
        return result;
    }

    private boolean isConfusingNumber(String num) {
        StringBuilder rotated = new StringBuilder();
        for (char c : num.toCharArray()) {
            rotated.append(CONFUSING_MAP.get(c));
        }
        return !num.equals(rotated.reverse().toString());
    }
}