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en/docs/chapter_heap/top_k.md
+33 -293
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@@ -78,15 +78,15 @@ Example code is as follows:
```python title="top_k.py"
def top_k_heap(nums: list[int], k: int) -> list[int]:
"""基于堆查找数组中最大的 k 个元素"""
# 初始化小顶堆
"""Using heap to find the largest k elements in an array"""
# Initialize min-heap
heap = []
# 将数组的前 k 个元素入堆
# Enter the first k elements of the array into the heap
for i in range(k):
heapq.heappush(heap, nums[i])
# 从第 k+1 个元素开始，保持堆的长度为 k
# From the k+1th element, keep the heap length as k
for i in range(k, len(nums)):
# 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
# If the current element is larger than the heap top element, remove the heap top element and enter the current element into the heap
if nums[i] > heap[0]:
heapq.heappop(heap)
heapq.heappush(heap, nums[i])
@@ -96,40 +96,23 @@ Example code is as follows:
=== "C++"
```cpp title="top_k.cpp"
/* 基于堆查找数组中最大的 k 个元素 */
priority_queue<int, vector<int>, greater<int>> topKHeap(vector<int> &nums, int k) {
// 初始化小顶堆
priority_queue<int, vector<int>, greater<int>> heap;
// 将数组的前 k 个元素入堆
for (int i = 0; i < k; i++) {
heap.push(nums[i]);
}
// 从第 k+1 个元素开始，保持堆的长度为 k
for (int i = k; i < nums.size(); i++) {
// 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
if (nums[i] > heap.top()) {
heap.pop();
heap.push(nums[i]);
}
}
return heap;
}
[class]{}-[func]{topKHeap}
```
=== "Java"
```java title="top_k.java"
/* 基于堆查找数组中最大的 k 个元素 */
/* Using heap to find the largest k elements in an array */
Queue<Integer> topKHeap(int[] nums, int k) {
// 初始化小顶堆
// Initialize min-heap
Queue<Integer> heap = new PriorityQueue<Integer>();
// 将数组的前 k 个元素入堆
// Enter the first k elements of the array into the heap
for (int i = 0; i < k; i++) {
heap.offer(nums[i]);
}
// 从第 k+1 个元素开始，保持堆的长度为 k
// From the k+1th element, keep the heap length as k
for (int i = k; i < nums.length; i++) {
// 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
// If the current element is larger than the heap top element, remove the heap top element and enter the current element into the heap
if (nums[i] > heap.peek()) {
heap.poll();
heap.offer(nums[i]);
@@ -142,323 +125,85 @@ Example code is as follows:
=== "C#"
```csharp title="top_k.cs"
/* 基于堆查找数组中最大的 k 个元素 */
PriorityQueue<int, int> TopKHeap(int[] nums, int k) {
// 初始化小顶堆
PriorityQueue<int, int> heap = new();
// 将数组的前 k 个元素入堆
for (int i = 0; i < k; i++) {
heap.Enqueue(nums[i], nums[i]);
}
// 从第 k+1 个元素开始，保持堆的长度为 k
for (int i = k; i < nums.Length; i++) {
// 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
if (nums[i] > heap.Peek()) {
heap.Dequeue();
heap.Enqueue(nums[i], nums[i]);
}
}
return heap;
}
[class]{top_k}-[func]{TopKHeap}
```
=== "Go"
```go title="top_k.go"
/* 基于堆查找数组中最大的 k 个元素 */
func topKHeap(nums []int, k int) *minHeap {
// 初始化小顶堆
h := &minHeap{}
heap.Init(h)
// 将数组的前 k 个元素入堆
for i := 0; i < k; i++ {
heap.Push(h, nums[i])
}
// 从第 k+1 个元素开始，保持堆的长度为 k
for i := k; i < len(nums); i++ {
// 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
if nums[i] > h.Top().(int) {
heap.Pop(h)
heap.Push(h, nums[i])
}
}
return h
}
[class]{}-[func]{topKHeap}
```
=== "Swift"
```swift title="top_k.swift"
/* 基于堆查找数组中最大的 k 个元素 */
func topKHeap(nums: [Int], k: Int) -> [Int] {
// 初始化一个小顶堆，并将前 k 个元素建堆
var heap = Heap(nums.prefix(k))
// 从第 k+1 个元素开始，保持堆的长度为 k
for i in nums.indices.dropFirst(k) {
// 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
if nums[i] > heap.min()! {
_ = heap.removeMin()
heap.insert(nums[i])
}
}
return heap.unordered
}
[class]{}-[func]{topKHeap}
```
=== "JS"
```javascript title="top_k.js"
/* 元素入堆 */
function pushMinHeap(maxHeap, val) {
// 元素取反
maxHeap.push(-val);
}
[class]{}-[func]{pushMinHeap}
/* 元素出堆 */
function popMinHeap(maxHeap) {
// 元素取反
return -maxHeap.pop();
}
[class]{}-[func]{popMinHeap}
/* 访问堆顶元素 */
function peekMinHeap(maxHeap) {
// 元素取反
return -maxHeap.peek();
}
[class]{}-[func]{peekMinHeap}
/* 取出堆中元素 */
function getMinHeap(maxHeap) {
// 元素取反
return maxHeap.getMaxHeap().map((num) => -num);
}
[class]{}-[func]{getMinHeap}
/* 基于堆查找数组中最大的 k 个元素 */
function topKHeap(nums, k) {
// 初始化小顶堆
// 请注意：我们将堆中所有元素取反，从而用大顶堆来模拟小顶堆
const maxHeap = new MaxHeap([]);
// 将数组的前 k 个元素入堆
for (let i = 0; i < k; i++) {
pushMinHeap(maxHeap, nums[i]);
}
// 从第 k+1 个元素开始，保持堆的长度为 k
for (let i = k; i < nums.length; i++) {
// 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
if (nums[i] > peekMinHeap(maxHeap)) {
popMinHeap(maxHeap);
pushMinHeap(maxHeap, nums[i]);
}
}
// 返回堆中元素
return getMinHeap(maxHeap);
}
[class]{}-[func]{topKHeap}
```
=== "TS"
```typescript title="top_k.ts"
/* 元素入堆 */
function pushMinHeap(maxHeap: MaxHeap, val: number): void {
// 元素取反
maxHeap.push(-val);
}
[class]{}-[func]{pushMinHeap}
/* 元素出堆 */
function popMinHeap(maxHeap: MaxHeap): number {
// 元素取反
return -maxHeap.pop();
}
[class]{}-[func]{popMinHeap}
/* 访问堆顶元素 */
function peekMinHeap(maxHeap: MaxHeap): number {
// 元素取反
return -maxHeap.peek();
}
[class]{}-[func]{peekMinHeap}
/* 取出堆中元素 */
function getMinHeap(maxHeap: MaxHeap): number[] {
// 元素取反
return maxHeap.getMaxHeap().map((num: number) => -num);
}
[class]{}-[func]{getMinHeap}
/* 基于堆查找数组中最大的 k 个元素 */
function topKHeap(nums: number[], k: number): number[] {
// 初始化小顶堆
// 请注意：我们将堆中所有元素取反，从而用大顶堆来模拟小顶堆
const maxHeap = new MaxHeap([]);
// 将数组的前 k 个元素入堆
for (let i = 0; i < k; i++) {
pushMinHeap(maxHeap, nums[i]);
}
// 从第 k+1 个元素开始，保持堆的长度为 k
for (let i = k; i < nums.length; i++) {
// 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
if (nums[i] > peekMinHeap(maxHeap)) {
popMinHeap(maxHeap);
pushMinHeap(maxHeap, nums[i]);
}
}
// 返回堆中元素
return getMinHeap(maxHeap);
}
[class]{}-[func]{topKHeap}
```
=== "Dart"
```dart title="top_k.dart"
/* 基于堆查找数组中最大的 k 个元素 */
MinHeap topKHeap(List<int> nums, int k) {
// 初始化小顶堆，将数组的前 k 个元素入堆
MinHeap heap = MinHeap(nums.sublist(0, k));
// 从第 k+1 个元素开始，保持堆的长度为 k
for (int i = k; i < nums.length; i++) {
// 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
if (nums[i] > heap.peek()) {
heap.pop();
heap.push(nums[i]);
}
}
return heap;
}
[class]{}-[func]{topKHeap}
```
=== "Rust"
```rust title="top_k.rs"
/* 基于堆查找数组中最大的 k 个元素 */
fn top_k_heap(nums: Vec<i32>, k: usize) -> BinaryHeap<Reverse<i32>> {
// BinaryHeap 是大顶堆，使用 Reverse 将元素取反，从而实现小顶堆
let mut heap = BinaryHeap::<Reverse<i32>>::new();
// 将数组的前 k 个元素入堆
for &num in nums.iter().take(k) {
heap.push(Reverse(num));
}
// 从第 k+1 个元素开始，保持堆的长度为 k
for &num in nums.iter().skip(k) {
// 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
if num > heap.peek().unwrap().0 {
heap.pop();
heap.push(Reverse(num));
}
}
heap
}
[class]{}-[func]{top_k_heap}
```
=== "C"
```c title="top_k.c"
/* 元素入堆 */
void pushMinHeap(MaxHeap *maxHeap, int val) {
// 元素取反
push(maxHeap, -val);
}
[class]{}-[func]{pushMinHeap}
/* 元素出堆 */
int popMinHeap(MaxHeap *maxHeap) {
// 元素取反
return -pop(maxHeap);
}
[class]{}-[func]{popMinHeap}
/* 访问堆顶元素 */
int peekMinHeap(MaxHeap *maxHeap) {
// 元素取反
return -peek(maxHeap);
}
[class]{}-[func]{peekMinHeap}
/* 取出堆中元素 */
int *getMinHeap(MaxHeap *maxHeap) {
// 将堆中所有元素取反并存入 res 数组
int *res = (int *)malloc(maxHeap->size * sizeof(int));
for (int i = 0; i < maxHeap->size; i++) {
res[i] = -maxHeap->data[i];
}
return res;
}
[class]{}-[func]{getMinHeap}
/* 取出堆中元素 */
int *getMinHeap(MaxHeap *maxHeap) {
// 将堆中所有元素取反并存入 res 数组
int *res = (int *)malloc(maxHeap->size * sizeof(int));
for (int i = 0; i < maxHeap->size; i++) {
res[i] = -maxHeap->data[i];
}
return res;
}
// 基于堆查找数组中最大的 k 个元素的函数
int *topKHeap(int *nums, int sizeNums, int k) {
// 初始化小顶堆
// 请注意：我们将堆中所有元素取反，从而用大顶堆来模拟小顶堆
int *empty = (int *)malloc(0);
MaxHeap *maxHeap = newMaxHeap(empty, 0);
// 将数组的前 k 个元素入堆
for (int i = 0; i < k; i++) {
pushMinHeap(maxHeap, nums[i]);
}
// 从第 k+1 个元素开始，保持堆的长度为 k
for (int i = k; i < sizeNums; i++) {
// 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
if (nums[i] > peekMinHeap(maxHeap)) {
popMinHeap(maxHeap);
pushMinHeap(maxHeap, nums[i]);
}
}
int *res = getMinHeap(maxHeap);
// 释放内存
delMaxHeap(maxHeap);
return res;
}
[class]{}-[func]{topKHeap}
```
=== "Kotlin"
```kotlin title="top_k.kt"
/* 基于堆查找数组中最大的 k 个元素 */
fun topKHeap(nums: IntArray, k: Int): Queue<Int> {
// 初始化小顶堆
val heap = PriorityQueue<Int>()
// 将数组的前 k 个元素入堆
for (i in 0..<k) {
heap.offer(nums[i])
}
// 从第 k+1 个元素开始，保持堆的长度为 k
for (i in k..<nums.size) {
// 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
if (nums[i] > heap.peek()) {
heap.poll()
heap.offer(nums[i])
}
}
return heap
}
[class]{}-[func]{topKHeap}
```
=== "Ruby"
```ruby title="top_k.rb"
### 基于堆查找数组中最大的 k 个元素 ###
def top_k_heap(nums, k)
# 初始化小顶堆
# 请注意：我们将堆中所有元素取反，从而用大顶堆来模拟小顶堆
max_heap = MaxHeap.new([])
# 将数组的前 k 个元素入堆
for i in 0...k
push_min_heap(max_heap, nums[i])
end
# 从第 k+1 个元素开始，保持堆的长度为 k
for i in k...nums.length
# 若当前元素大于堆顶元素，则将堆顶元素出堆、当前元素入堆
if nums[i] > peek_min_heap(max_heap)
pop_min_heap(max_heap)
push_min_heap(max_heap, nums[i])
end
end
get_min_heap(max_heap)
end
[class]{}-[func]{top_k_heap}
```
=== "Zig"
@@ -467,11 +212,6 @@ Example code is as follows:
[class]{}-[func]{topKHeap}
```
??? pythontutor "Code Visualization"
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=import%20heapq%0A%0Adef%20top_k_heap%28nums%3A%20list%5Bint%5D,%20k%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E5%9F%BA%E4%BA%8E%E5%A0%86%E6%9F%A5%E6%89%BE%E6%95%B0%E7%BB%84%E4%B8%AD%E6%9C%80%E5%A4%A7%E7%9A%84%20k%20%E4%B8%AA%E5%85%83%E7%B4%A0%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%B0%8F%E9%A1%B6%E5%A0%86%0A%20%20%20%20heap%20%3D%20%5B%5D%0A%20%20%20%20%23%20%E5%B0%86%E6%95%B0%E7%BB%84%E7%9A%84%E5%89%8D%20k%20%E4%B8%AA%E5%85%83%E7%B4%A0%E5%85%A5%E5%A0%86%0A%20%20%20%20for%20i%20in%20range%28k%29%3A%0A%20%20%20%20%20%20%20%20heapq.heappush%28heap,%20nums%5Bi%5D%29%0A%20%20%20%20%23%20%E4%BB%8E%E7%AC%AC%20k%2B1%20%E4%B8%AA%E5%85%83%E7%B4%A0%E5%BC%80%E5%A7%8B%EF%BC%8C%E4%BF%9D%E6%8C%81%E5%A0%86%E7%9A%84%E9%95%BF%E5%BA%A6%E4%B8%BA%20k%0A%20%20%20%20for%20i%20in%20range%28k,%20len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E5%BD%93%E5%89%8D%E5%85%83%E7%B4%A0%E5%A4%A7%E4%BA%8E%E5%A0%86%E9%A1%B6%E5%85%83%E7%B4%A0%EF%BC%8C%E5%88%99%E5%B0%86%E5%A0%86%E9%A1%B6%E5%85%83%E7%B4%A0%E5%87%BA%E5%A0%86%E3%80%81%E5%BD%93%E5%89%8D%E5%85%83%E7%B4%A0%E5%85%A5%E5%A0%86%0A%20%20%20%20%20%20%20%20if%20nums%5Bi%5D%20%3E%20heap%5B0%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20heapq.heappop%28heap%29%0A%20%20%20%20%20%20%20%20%20%20%20%20heapq.heappush%28heap,%20nums%5Bi%5D%29%0A%20%20%20%20return%20heap%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B1,%207,%206,%203,%202%5D%0A%20%20%20%20k%20%3D%203%0A%0A%20%20%20%20res%20%3D%20top_k_heap%28nums,%20k%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=import%20heapq%0A%0Adef%20top_k_heap%28nums%3A%20list%5Bint%5D,%20k%3A%20int%29%20-%3E%20list%5Bint%5D%3A%0A%20%20%20%20%22%22%22%E5%9F%BA%E4%BA%8E%E5%A0%86%E6%9F%A5%E6%89%BE%E6%95%B0%E7%BB%84%E4%B8%AD%E6%9C%80%E5%A4%A7%E7%9A%84%20k%20%E4%B8%AA%E5%85%83%E7%B4%A0%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%B0%8F%E9%A1%B6%E5%A0%86%0A%20%20%20%20heap%20%3D%20%5B%5D%0A%20%20%20%20%23%20%E5%B0%86%E6%95%B0%E7%BB%84%E7%9A%84%E5%89%8D%20k%20%E4%B8%AA%E5%85%83%E7%B4%A0%E5%85%A5%E5%A0%86%0A%20%20%20%20for%20i%20in%20range%28k%29%3A%0A%20%20%20%20%20%20%20%20heapq.heappush%28heap,%20nums%5Bi%5D%29%0A%20%20%20%20%23%20%E4%BB%8E%E7%AC%AC%20k%2B1%20%E4%B8%AA%E5%85%83%E7%B4%A0%E5%BC%80%E5%A7%8B%EF%BC%8C%E4%BF%9D%E6%8C%81%E5%A0%86%E7%9A%84%E9%95%BF%E5%BA%A6%E4%B8%BA%20k%0A%20%20%20%20for%20i%20in%20range%28k,%20len%28nums%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E5%BD%93%E5%89%8D%E5%85%83%E7%B4%A0%E5%A4%A7%E4%BA%8E%E5%A0%86%E9%A1%B6%E5%85%83%E7%B4%A0%EF%BC%8C%E5%88%99%E5%B0%86%E5%A0%86%E9%A1%B6%E5%85%83%E7%B4%A0%E5%87%BA%E5%A0%86%E3%80%81%E5%BD%93%E5%89%8D%E5%85%83%E7%B4%A0%E5%85%A5%E5%A0%86%0A%20%20%20%20%20%20%20%20if%20nums%5Bi%5D%20%3E%20heap%5B0%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20heapq.heappop%28heap%29%0A%20%20%20%20%20%20%20%20%20%20%20%20heapq.heappush%28heap,%20nums%5Bi%5D%29%0A%20%20%20%20return%20heap%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B1,%207,%206,%203,%202%5D%0A%20%20%20%20k%20%3D%203%0A%0A%20%20%20%20res%20%3D%20top_k_heap%28nums,%20k%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
A total of $n$ rounds of heap insertions and deletions are performed, with the maximum heap size being $k$, hence the time complexity is $O(n \log k)$. This method is very efficient; when $k$ is small, the time complexity tends towards $O(n)$; when $k$ is large, the time complexity will not exceed $O(n \log n)$.
Additionally, this method is suitable for scenarios with dynamic data streams. By continuously adding data, we can maintain the elements within the heap, thereby achieving dynamic updates of the largest $k$ elements.