Problem Given an integer array, heapify it into a min-heap array. For a heap array A, A[0] is the root of heap, and for each A[i], A[i * 2 + 1] is the left child of A[i] and A[i * 2 + 2] is the right ...
...下: HeapSort(A) BuildHeap(A) for i = n-1 to 0 swap(A[0],A[i]) n = n - 1 Heapify(A, 0) BuildHeap(A) n = elemens_in(A) for i = floor(n / 2) to 0 Heapify(A, i) Heapify(A, i) left = 2i+1; right = 2i +...
... arr[0] = arr[i]; arr[i] = temp; // call max heapify on the reduced heap heapify(arr, i, 0); } } // To heapify a subtree rooted with node...
... = [...arr]; this.size = this.data.length; } } max-heapify最大堆操作 max-heapify是把每一個(gè)不滿足最大堆性質(zhì)的分支節(jié)點(diǎn)進(jìn)行調(diào)整的一個(gè)操作。 如上圖: 調(diào)整分支節(jié)點(diǎn)2(分支節(jié)點(diǎn)2不滿足最大堆的性質(zhì)) 默認(rèn)該分支節(jié)點(diǎn)...
...h.floor(len(arr)/2),-1,-1):#構(gòu)建堆由下往上構(gòu)建所以用-1 heapify(arr,i) def heapify(arr, i): left = 2*i+1 right = 2*i+2 largest = i if left < arrLen and arr[left] > arr[largest]: ...
... $t = $x; $x = $y; $y = $t; } public function max_heapify(&$arr,$start,$end){ $dad = $start; $son = $dad*2+1; if($son >=$end) return; ...
...現(xiàn)。 heapq 的使用 創(chuàng)建堆有兩個(gè)基本的方法:heappush() 和 heapify(),取出堆頂元素用 heappop()。 heappush() 是用來向已有的堆中添加元素,一般從空列表開始構(gòu)建: import heapq data = [97, 38, 27, 50, 76, 65, 49, 13] heap = [] for n in data: heapq.he....
...[size - 1]; data[-- this.size] = 0; //進(jìn)行堆化 heapify(data, size, 0); return result; } //堆化函數(shù) private void heapify(int[] data, int size, int i){ ...
...介紹。heapq(Python內(nèi)置的模塊) __all__ = [heappush, heappop, heapify, heapreplace, merge, nlargest, nsmallest, heappushpop] 接下來我們一一介紹。nlargest與nsmallest,通過字面意思可以看出方法大致的作用,接下來動(dòng)手測(cè)驗(yàn) nlargest(n, ite....
...操作跟最小堆類一樣,這里就不多加贅述。 堆排序算法 heapify(array) { if (array) { this.heap = array; } const maxIndex = Math.floor(this.size() / 2) - 1; for (let i = 0; i = 0; i -= 1){ heapify(ar...
... while (left.length && right.length) { if (left[0] = 0; i--) { heapify(arr, i); } } function heapify(arr, i) { // 堆調(diào)整 var left = 2 * i + 1, right = 2 * i + 2, ...
...le (left.length && right.length) { if (left[0] = 0; i--) { heapify(arr, i); } } function heapify(arr, i) { //堆調(diào)整 var left = 2 * i + 1, right = 2 * i + 2, ...
... for (let i = Math.floor(array.length / 2 - 1); i >= 0; i--) { heapify(array, i, array.length); } // 排序,每一次 for 循環(huán)找出一個(gè)當(dāng)前最大值,數(shù)組長度減一 for (let i = Math.floor(array.length - 1);...
...就是說每次pop的時(shí)候取出的是最小的元素 首先使用heapq.heapify將一個(gè)列表初始化為堆 >>> import heapq >>> l = [-1, 2, 5, 0, 8] >>> heapq.heapify(l) >>> print(l) [-1, 0, 5, 2, 8] 然后就可以調(diào)用heapq.heappush和heapq.heappop對(duì)堆進(jìn)行增加和刪除操作了 >>> hea...
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