feat: add typescript solution to lc problem: No.2044 (#586)

No.2044.Count Number of Maximum Bitwise-OR Subsets
doocs · Oct 19, 2021 · 6ab3755 · 6ab3755
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diff --git a/solution/2000-2099/2044.Count Number of Maximum Bitwise-OR Subsets/README.md b/solution/2000-2099/2044.Count Number of Maximum Bitwise-OR Subsets/README.md
@@ -117,6 +117,29 @@ class Solution {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function countMaxOrSubsets(nums: number[]): number {
+    let n = nums.length;
+    let max = 0;
+    for (let i = 0; i < n; i++) {
+        max |= nums[i];
+    }
+    let ans = 0;
+    function dfs (pre: number, depth: number): void {
+        if (depth == n) {
+            if (pre == max) ++ans;
+            return;
+        }
+        dfs(pre, depth + 1);
+        dfs(pre | nums[depth], depth + 1);
+    }
+    dfs(0, 0);
+    return ans;
+};
+```
+
 ### **C++**
 
 ```cpp

diff --git a/solution/2000-2099/2044.Count Number of Maximum Bitwise-OR Subsets/README_EN.md b/solution/2000-2099/2044.Count Number of Maximum Bitwise-OR Subsets/README_EN.md
@@ -107,6 +107,29 @@ class Solution {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function countMaxOrSubsets(nums: number[]): number {
+    let n = nums.length;
+    let max = 0;
+    for (let i = 0; i < n; i++) {
+        max |= nums[i];
+    }
+    let ans = 0;
+    function dfs (pre: number, depth: number): void {
+        if (depth == n) {
+            if (pre == max) ++ans;
+            return;
+        }
+        dfs(pre, depth + 1);
+        dfs(pre | nums[depth], depth + 1);
+    }
+    dfs(0, 0);
+    return ans;
+};
+```
+
 ### **C++**
 
 ```cpp

diff --git a/solution/2000-2099/2044.Count Number of Maximum Bitwise-OR Subsets/Solution.ts b/solution/2000-2099/2044.Count Number of Maximum Bitwise-OR Subsets/Solution.ts
@@ -0,0 +1,18 @@
+function countMaxOrSubsets(nums: number[]): number {
+    let n = nums.length;
+    let max = 0;
+    for (let i = 0; i < n; i++) {
+        max |= nums[i];
+    }
+    let ans = 0;
+    function dfs (pre: number, depth: number): void {
+        if (depth == n) {
+            if (pre == max) ++ans;
+            return;
+        }
+        dfs(pre, depth + 1);
+        dfs(pre | nums[depth], depth + 1);
+    }
+    dfs(0, 0);
+    return ans;
+};