https://leetcode.cn/problems/boolean-evaluation-lcci/

Author: masx200

README.md

@@ -10,6 +10,8 @@ leetcode 测试
 
 ##### 包含的内容如下
 
+https://leetcode.cn/problems/boolean-evaluation-lcci/
+
 https://leetcode.cn/problems/build-an-array-with-stack-operations/
 
 https://leetcode.cn/problems/using-a-robot-to-print-the-lexicographically-smallest-string/

boolean-evaluation-lcci/index.ts

@@ -0,0 +1,42 @@
+function countEval(s: string, result: number): number {
+    const n = s.length;
+    // 创建三维数组
+    const dp: number[][][] = new Array(n)
+        .fill(0)
+        .map(() => new Array(n).fill(0).map(() => new Array(2).fill(0)));
+    // 初始化basecase s[i] 为 1,则 dp[i][i][1] = 1, dp[i][i][0] = 0
+    for (let i = 0; i < n; i += 2) {
+        let tmp = 0;
+        if (s[i] == "1") tmp = 1;
+        dp[i][i][0] = 1 - tmp;
+        dp[i][i][1] = tmp;
+    }
+
+    // 从0开始，每次+2
+    for (let step = 0; step < n; step += 2) {
+        // 计算 i - （i+step） 的情况， j 为中间间隔点 s[j] 为符号
+        for (let i = 0; i + step < n; i += 2) {
+            for (let j = i + 1; j < i + step; j += 2) {
+                const left0 = dp[i][j - 1][0],
+                    left1 = dp[i][j - 1][1];
+                const right0 = dp[j + 1][i + step][0],
+                    right1 = dp[j + 1][i + step][1];
+                if (s[j] == "&") {
+                    dp[i][i + step][0] += left0 * (right0 + right1) +
+                        left1 * right0;
+                    dp[i][i + step][1] += left1 * right1;
+                } else if (s[j] == "|") {
+                    dp[i][i + step][0] += left0 * right0;
+                    dp[i][i + step][1] += left0 * right1 +
+                        left1 * (right0 + right1);
+                } else {
+                    //s[j]=='^'
+                    dp[i][i + step][0] += left0 * right0 + left1 * right1;
+                    dp[i][i + step][1] += left0 * right1 + left1 * right0;
+                }
+            }
+        }
+    }
+    return dp[0][n - 1][result];
+}
+export default countEval;