diff --git a/src/main/java/leetcode/editor/cn/BooleanEvaluationLcci.java b/src/main/java/leetcode/editor/cn/BooleanEvaluationLcci.java
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+++ b/src/main/java/leetcode/editor/cn/BooleanEvaluationLcci.java
@@ -0,0 +1,76 @@
+//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)
+// 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。 
+//
+// 示例 1: 
+//
+// 输入: s = "1^0|0|1", result = 0
+//
+//输出: 2
+//解释: 两种可能的括号方法是
+//1^(0|(0|1))
+//1^((0|0)|1)
+// 
+//
+// 示例 2: 
+//
+// 输入: s = "0&0&0&1^1|0", result = 1
+//
+//输出: 10 
+//
+// 提示： 
+//
+// 
+// 运算符的数量不超过 19 个 
+// 
+// Related Topics 栈 字符串 
+// 👍 43 👎 0
+
+package leetcode.editor.cn;
+//面试题 08.14:布尔运算
+public class BooleanEvaluationLcci{
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new BooleanEvaluationLcci().new Solution();
+    }
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+class Solution {
+    public int countEval(String s, int result) {
+        if (s.length() == 0) {
+            return 0;
+        }
+        if (s.length() == 1) {
+            return (s.charAt(0) - '0') == result ? 1 : 0;
+        }
+        char[] ch = s.toCharArray();
+        int[][][] dp = new int[ch.length][ch.length][2];
+        for (int i = 0; i < ch.length; i++) {
+            if (ch[i] == '0' || ch[i] == '1') {
+                dp[i][i][ch[i] - '0'] = 1;
+            }
+        }
+        for (int len = 2; len <= ch.length; len += 2) {
+            for (int start = 0; start <= ch.length - len; start += 2) {
+                int end = start + len;
+                for (int k = start + 1; k <= end - 1; k += 2) {
+                    if (ch[k] == '&') {
+                        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];
+                        dp[start][end][1] += dp[start][k - 1][1] * dp[k + 1][end][1];
+                    }
+                    if (ch[k] == '|') {
+                        dp[start][end][0] += dp[start][k - 1][0] * dp[k + 1][end][0];
+                        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];
+                    }
+                    if (ch[k] == '^') {
+                        dp[start][end][0] += dp[start][k - 1][0] * dp[k + 1][end][0] + dp[start][k - 1][1] * dp[k + 1][end][1];
+                        dp[start][end][1] += dp[start][k - 1][1] * dp[k + 1][end][0] + dp[start][k - 1][0] * dp[k + 1][end][1];
+                    }
+                }
+            }
+        }
+        return dp[0][ch.length - 1][result];
+    }
+}
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
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diff --git a/src/main/java/leetcode/editor/cn/BooleanEvaluationLcci.md b/src/main/java/leetcode/editor/cn/BooleanEvaluationLcci.md
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+++ b/src/main/java/leetcode/editor/cn/BooleanEvaluationLcci.md
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+<p>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 <code>0</code> (false)、<code>1</code> (true)、<code>&amp;</code> (AND)、 <code>|</code> (OR) 和 <code>^</code> (XOR) 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。</p>
+
+<p><strong>示例 1:</strong></p>
+
+<pre><strong>输入: </strong>s = &quot;1^0|0|1&quot;, result = 0
+
+<strong>输出: </strong>2
+<strong>解释:</strong>&nbsp;两种可能的括号方法是
+1^(0|(0|1))
+1^((0|0)|1)
+</pre>
+
+<p><strong>示例 2:</strong></p>
+
+<pre><strong>输入: </strong>s = &quot;0&amp;0&amp;0&amp;1^1|0&quot;, result = 1
+
+<strong>输出: </strong>10</pre>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li>运算符的数量不超过 19 个</li>
+</ul>
+<div><div>Related Topics</div><div><li>栈</li><li>字符串</li></div></div>\n<div><li>👍 43</li><li>👎 0</li></div>
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