feat: add 3rd ts solution to lc problem: No.0347

Author: rain84

solution/0300-0399/0347.Top K Frequent Elements/README.md

@@ -282,4 +282,39 @@ function topKFrequent(nums: number[], k: number): number[] {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### Solution 3. Priority queue
+
+<!-- tabs:start -->
+
+#### TypeScript
+
+```ts
+function topKFrequent(nums: number[], k: number): number[] {
+    const ans = Array<number>(k);
+    const cnt = new Map<number, number>();
+    const minPQ = new MinPriorityQueue();
+
+    for (const x of nums) {
+        cnt.set(x, (cnt.get(x) ?? 0) + 1);
+    }
+
+    for (const [x, c] of cnt) {
+        minPQ.enqueue(x, c);
+        if (minPQ.size() > k) minPQ.dequeue();
+    }
+
+    while (k--) {
+        ans[k] = minPQ.dequeue().element;
+    }
+
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->

solution/0300-0399/0347.Top K Frequent Elements/README_EN.md

@@ -270,4 +270,39 @@ function topKFrequent(nums: number[], k: number): number[] {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### Solution 3. Priority queue
+
+<!-- tabs:start -->
+
+#### TypeScript
+
+```ts
+function topKFrequent(nums: number[], k: number): number[] {
+    const ans = Array<number>(k);
+    const cnt = new Map<number, number>();
+    const minPQ = new MinPriorityQueue();
+
+    for (const x of nums) {
+        cnt.set(x, (cnt.get(x) ?? 0) + 1);
+    }
+
+    for (const [x, c] of cnt) {
+        minPQ.enqueue(x, c);
+        if (minPQ.size() > k) minPQ.dequeue();
+    }
+
+    while (k--) {
+        ans[k] = minPQ.dequeue().element;
+    }
+
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->

solution/0300-0399/0347.Top K Frequent Elements/Solution3.ts

@@ -0,0 +1,20 @@
+function topKFrequent(nums: number[], k: number): number[] {
+    const ans = Array<number>(k);
+    const cnt = new Map<number, number>();
+    const minPQ = new MinPriorityQueue();
+
+    for (const x of nums) {
+        cnt.set(x, (cnt.get(x) ?? 0) + 1);
+    }
+
+    for (const [x, c] of cnt) {
+        minPQ.enqueue(x, c);
+        if (minPQ.size() > k) minPQ.dequeue();
+    }
+
+    while (k--) {
+        ans[k] = minPQ.dequeue().element;
+    }
+
+    return ans;
+}