76 lines
2.7 KiB
Java
76 lines
2.7 KiB
Java
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//给定一个布尔表达式和一个期望的布尔结果 result，布尔表达式由 0 (false)、1 (true)、& (AND)、 | (OR) 和 ^ (XOR)
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// 符号组成。实现一个函数，算出有几种可使该表达式得出 result 值的括号方法。
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//
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// 示例 1:
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//
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// 输入: s = "1^0|0|1", result = 0
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//
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//输出: 2
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//解释: 两种可能的括号方法是
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//1^(0|(0|1))
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//1^((0|0)|1)
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//
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//
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// 示例 2:
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//
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// 输入: s = "0&0&0&1^1|0", result = 1
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//
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//输出: 10
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//
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// 提示：
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//
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//
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// 运算符的数量不超过 19 个
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//
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// Related Topics 栈 字符串
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// 👍 43 👎 0
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package leetcode.editor.cn;
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//面试题 08.14:布尔运算
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public class BooleanEvaluationLcci{
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public static void main(String[] args) {
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//测试代码
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Solution solution = new BooleanEvaluationLcci().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public int countEval(String s, int result) {
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if (s.length() == 0) {
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return 0;
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}
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if (s.length() == 1) {
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return (s.charAt(0) - '0') == result ? 1 : 0;
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}
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char[] ch = s.toCharArray();
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int[][][] dp = new int[ch.length][ch.length][2];
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for (int i = 0; i < ch.length; i++) {
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if (ch[i] == '0' || ch[i] == '1') {
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dp[i][i][ch[i] - '0'] = 1;
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}
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}
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for (int len = 2; len <= ch.length; len += 2) {
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for (int start = 0; start <= ch.length - len; start += 2) {
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int end = start + len;
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for (int k = start + 1; k <= end - 1; k += 2) {
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if (ch[k] == '&') {
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dp[start][end][0] += dp[start][k - 1][0] * dp[k + 1][end][0] + dp[start][k - 1][0] * dp[k + 1][end][1] + dp[start][k - 1][1] * dp[k + 1][end][0];
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dp[start][end][1] += dp[start][k - 1][1] * dp[k + 1][end][1];
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}
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if (ch[k] == '|') {
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dp[start][end][0] += dp[start][k - 1][0] * dp[k + 1][end][0];
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dp[start][end][1] += dp[start][k - 1][0] * dp[k + 1][end][1] + dp[start][k - 1][1] * dp[k + 1][end][0] + dp[start][k - 1][1] * dp[k + 1][end][1];
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}
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if (ch[k] == '^') {
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dp[start][end][0] += dp[start][k - 1][0] * dp[k + 1][end][0] + dp[start][k - 1][1] * dp[k + 1][end][1];
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dp[start][end][1] += dp[start][k - 1][1] * dp[k + 1][end][0] + dp[start][k - 1][0] * dp[k + 1][end][1];
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}
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}
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}
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}
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return dp[0][ch.length - 1][result];
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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