9.11 given a Boolean expression, it consists of 0, 1, &, |, ^, and other symbols, as well as a desired boolean result to implement a function, this expression can be used to obtain the result value by calculating the placement of several parentheses.

 

Idea: recursive solution of enumeration Split points, which can be optimized by DP.

Pay attention to recursive termination conditions.

Note ^ & | there are three different statistical conditions.

 

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));    }}

 

9.11 given a Boolean expression, it consists of 0, 1, &, |, ^, and other symbols, as well as a desired boolean result to implement a function, this expression can be used to obtain the result value by calculating the placement of several parentheses.