9.11 给定一个布尔表达式，由0、1、&、|、^等符号组成，以及一个想要的布尔结果result，实现一个函数，算出有几种括号的放法可使该表达式得出result的值。

import java.util.HashMap;
import java.util.Map;

public class Solution {

    int countR(String terms, boolean result, int start, int end, Map<String, Integer> cache) {
        String key = "" + result + start + end;
        if (cache.containsKey(key)) {
            return cache.get(key);
        }

        if (start == end) {
            if (terms.charAt(start) == ‘0‘) {
                if (result)
                    return 0;
                else
                    return 1;
            } else {
                if (result)
                    return 1;
                else
                    return 0;
            }

        }
        int count = 0;
        if (result) {
            for (int i = start + 1; i < end; i++) {
                char op = terms.charAt(i);
                if (op == ‘&‘) {
                    count += countR(terms, true, start, i - 1, cache) * countR(terms, true, i + 1, end, cache);
                } else if (op == ‘|‘) {
                    count += countR(terms, true, start, i - 1, cache) * countR(terms, true, i + 1, end, cache);
                    count += countR(terms, true, start, i - 1, cache) * countR(terms, false, i + 1, end, cache);
                    count += countR(terms, false, start, i - 1, cache) * countR(terms, true, i + 1, end, cache);

                } else if (op == ‘^‘) {
                    count += countR(terms, true, start, i - 1, cache) * countR(terms, false, i + 1, end, cache);
                    count += countR(terms, false, start, i - 1, cache) * countR(terms, true, i + 1, end, cache);
                }

            }

        } else {
            for (int i = start + 1; i < end; i++) {
                char op = terms.charAt(i);
                if (op == ‘&‘) {
                    count += countR(terms, true, start, i - 1, cache) * countR(terms, false, i + 1, end, cache);
                    count += countR(terms, false, start, i - 1, cache) * countR(terms, true, i + 1, end, cache);
                    count += countR(terms, false, start, i - 1, cache) * countR(terms, false, i + 1, end, cache);

                } else if (op == ‘|‘) {
                    count += countR(terms, false, start, i - 1, cache) * countR(terms, false, i + 1, end, cache);

                } else if (op == ‘^‘) {
                    count += countR(terms, true, start, i - 1, cache) * countR(terms, true, i + 1, end, cache);
                    count += countR(terms, false, start, i - 1, cache) * countR(terms, false, i + 1, end, cache);
                }

            }

        }
        cache.put(key, count);
        return count;
    }

    public static void main(String[] args) {
        String terms = "0^0|1&1^1|0|1";
        boolean result = true;
        Map<String, Integer> cache = new HashMap<String, Integer>();
        System.out.println(new Solution().countR(terms, result, 0, terms.length() - 1, cache));

    }
}