我写了一些C代码来分析构建堆和运行heapsort的比较和运行时的数量。但是，我不确定我的代码输出是否有意义。 Heapsort应该在O（n log n）执行，但我看到的比较数似乎并不是非常接近。例如，对于大小为n = 100的输入，我看到~200比较构建堆和堆排序中的~800比较。我只是分析数据错误，或者我在代码中收集比较的方式有问题吗？
我可以提供github的链接，如果它会对任何人产生影响。
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
void bottom_up_heap_sort(int*, int);
void heap_sort(int*, int);
void sift_up(int*, int);
void sift_down(int*, int);
void build_max_heap(int*, int);
void bottom_up_build_max_heap(int*, int);
void randomize_in_place(int*, int);
int* generate_array(int);
void swap(int*, int*);
int cmp(int, int);
void print_array(int*, int);
int heapsize;
unsigned long comparison_counter;
clock_t begin, end;
double time_spent;
int main() {
int k, N;
int* A;
int* B;
int i;
printf("Testing Sift_Down Heap Sort\n");
for(k = 2; k <= 5; k++) {
comparison_counter = 0;
N = (int)pow((double)10, k);
begin = clock();
A = generate_array(N);
end = clock();
time_spent = (double)(end - begin) / CLOCKS_PER_SEC;
printf("Time Spent Generating Array: %f\n", time_spent);
// print the first unsorted array
//printf("Unsorted Array:\n");
//print_array(A, N);
begin = clock();
// call heap_sort on the first unsorted array
heap_sort(A, N);
end = clock();
time_spent = (double)(end - begin) / CLOCKS_PER_SEC;
// show that the array is now sorted
//printf("Sorted array: \n");
//print_array(A, N);
printf("Done with k = %d\n", k);
printf("Comparisons for Heap Sort: %lu\n", comparison_counter);
printf("Time Spent on Heap Sort: %f\n", time_spent);
printf("\n");
}
printf("----------------------------------\n");
printf("Testing Sift_Up Heap Sort\n");
for(k = 2; k <= 5; k++) {
comparison_counter = 0;
N = (int)pow((double)10, k);
begin = clock();
B = generate_array(N);
end = clock();
time_spent = (double)(end - begin) / CLOCKS_PER_SEC;
printf("Time Spent Generating Array: %f\n", time_spent);
// print the unsorted array
//printf("Unsorted Array:\n");
//print_array(B, N);
begin = clock();
// call heap_sort on the unsorted array
bottom_up_heap_sort(B, N);
end = clock();
time_spent = (double)(end - begin) / CLOCKS_PER_SEC;
// show that the array is now sorted
//printf("Sorted array: \n");
//print_array(B, N);
printf("Done with k = %d\n", k);
printf("Comparisons for Heap Sort: %lu\n", comparison_counter);
printf("Time Spent on Heap Sort: %f\n", time_spent);
printf("\n");
}
printf("----------------------------------\n");
return 0;
}
void bottom_up_heap_sort(int* arr, int len) {
int i;
// build a max heap from the bottom up using sift up
bottom_up_build_max_heap(arr, len);
printf("Comparisons for heap construction: %lu\n", comparison_counter);
comparison_counter = 0;
for(i = len-1; i >= 0; i--) {
// swap the last leaf and the root
swap(&arr[i], &arr[0]);
// remove the already sorted values
len--;
// repair the heap
bottom_up_build_max_heap(arr, len);
}
}
void heap_sort(int* arr, int len) {
int i;
// build a max heap from the array
build_max_heap(arr, len);
printf("Comparisons for heap construction: %lu\n", comparison_counter);
comparison_counter = 0;
for(i = len-1; i >= 1; i--) {
swap(&arr[0], &arr[i]); // move arr[0] to its sorted place
// remove the already sorted values
heapsize--;
sift_down(arr, 0); // repair the heap
}
}
void sift_down(int* arr, int i) {
int c = 2*i+1;
int largest;
if(c >= heapsize) return;
// locate largest child of i
if((c+1 < heapsize) && cmp(arr[c+1], arr[c]) > 0) {
c++;
}
// if child is larger than i, swap them
if(cmp(arr[c], arr[i]) > 0) {
swap(&arr[c], &arr[i]);
sift_down(arr, c);
}
}
void sift_up(int* arr, int i) {
if(i == 0) return; // at the root
// if the current node is larger than its parent, swap them
if(cmp(arr[i], arr[(i-1)/2]) > 0) {
swap(&arr[i], &arr[(i-1)/2]);
// sift up to repair the heap
sift_up(arr, (i-1)/2);
}
}
void bottom_up_build_max_heap(int* arr, int len) {
int i;
for(i = 0; i < len; i++) {
sift_up(arr, i);
}
}
void build_max_heap(int* arr, int len) {
heapsize = len;
int i;
for(i = len/2; i >= 0; i--) {
// invariant: arr[k], i < k <= n are roots of proper heaps
sift_down(arr, i);
}
}
void randomize_in_place(int* arr, int n) {
int j, k;
double val;
time_t t;
// init the random number generator
srand((unsigned)time(&t));
// randomization code from class notes
for(j = 0; j < n-1; j++) {
val = ((double)random()) / 0x7FFFFFFF;
k = j + val*(n-j);
swap(&arr[k], &arr[j]);
}
}
// this function is responsible for creating and populating an array
// of size k, and randomizing the locations of its elements
int* generate_array(int k) {
int* arr = (int*) malloc(sizeof(int)*k-1);
int i, j, x, N;
double val;
time_t t;
// init the random number generator
srand((unsigned)time(&t));
// fill the array with values from 1..N
for(i = 0; i <= k-1; i++) {
arr[i] = i+1;
}
N = (int)pow((double)10, 5);
// randomize the elements of the array for 10^5 iterations
for(i = 0; i < N; i++) {
randomize_in_place(arr, k);
}
return arr;
}
// swap two elements
void swap(int* a, int* b) {
int temp = *a;
*a = *b;
*b = temp;
}
int cmp(int a, int b) {
comparison_counter++;
if(a > b) return 1;
else if(a < b) return -1;
else return 0;
}
// print out an array by iterating through
void print_array(int* arr, int size) {
int i;
for(i = 0; i < size; i++) {
printf("%d ", arr[i]);
}
}
答案 0 :(得分：1)
这种小n值的实际数字并不重要，因为复杂性中省略了常数因子。重要的是算法的增长，测量越来越大的n值，并绘制它们应该给出与理论复杂度大致相同的图形。
我尝试了几个n的代码，复杂性的增加大约为O（n logn）
答案 1 :(得分：1)
O(n log n)（或一般O(f(x))）并不能让您了解单点的预期价值。
这是因为大O符号忽略了常数因素。换句话说，n * log(n)，0.000001 * n * log(n)和1000000 * n * log(n)都位于O(n log n)。因此，n的特定值的结果完全未确定。
你可以从big-O表示法中推断出修改控制变量的效果。如果函数涉及O(n)个操作，那么预计加倍n会使操作数增加一倍。如果函数涉及O(n2)个操作，那么预计加倍n将使操作数量翻两番。等等。