英文:
Determine Big-O of heap creation function
问题
以下是翻译好的部分：
这个 heapify 方法的时间复杂度的大O表示是多少？我原以为是 O(log n)，但好像不太对？如何使它变成 O(log n)？
假设我们有以下代码：
int arr[] = {1, 3, 5, 2, 7, 9, 4, 6};int n = arr.length;int i = 0; // 起始位置void heapify(int arr[], int n, int i) {int largest = i;int left = 2 * i + 1;int right = 2 * i + 2;if (left < n && arr[left] > arr[largest]) {largest = left;}if (right < n && arr[right] > arr[largest]) {largest = right;}if (largest != i) {int swap = arr[i];arr[i] = arr[largest];arr[largest] = swap;heapify(arr, n, largest);}}
英文:
What is the Big-O notation of this heapify method? I was expecting O(log n) but it does not add up? How do I make it O(log n)?
Assuming we have this:
int arr[] = {1, 3, 5, 2, 7, 9, 4, 6};int n = arr.length;int i = 0; // Start positionvoid heapify(int arr[], int n, int i) {int largest = i;int left = 2 * i + 1;int right = 2 * i + 2;if(left < n && arr[left] > arr[largest]) {largest = left;}if(right < n && arr[right] > arr[largest]) {largest = right;}if(largest != i) {int swap = arr[i];arr[i] = arr[largest];arr[largest] = swap;heapify(arr, n, largest);}}
Thanks,
答案1
得分: 1
这已经是 O(log n)。i 的值在每次递归调用时至少加倍，当 i >= n 时停止，在第一次调用后，i 至少为 1，因此递归调用最多可以有 O(log n) 次。
英文:
This is already O(log n). The value of i at least doubles with every recursive call, and when i >= n it stops, and after the first call i is at least one, so there can be at most O(log n) recursive calls.
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