网站改版设计外包网站有哪些
目录
一 降序(建小堆)
二 升序 (建大堆)
三 优化(以升序为例)
四 TOP-K问题
一 降序(建小堆)
void Swap(int* x, int* y)
{int tmp = *x;*x = *y;*y = tmp;
}//降序 建小堆
void AdjustUp(int* a, int child)
{int parent = (child - 1) / 2;while (child > 0){if (a[child] < a[parent]){Swap(&a[child], &a[parent]);child = parent;parent = (child - 1) / 2;}else{break;}}
}void AdjustDown(int* a, int n, int parent)
{int child = parent * 2 + 1;while (child < n){if (child + 1 < n && a[child + 1] < a[child]){child++;}if (a[child] < a[parent]){Swap(&a[child], &a[parent]);parent = child;child = parent * 2 + 1;}else{break;}}
}void HeapSort(int* a, int n)
{int end = n - 1;int i = 0;//建堆for (i = 1; i < n; i++){AdjustUp(a, i);}while (end > 0){Swap(&a[0], &a[end]);AdjustDown(a, end, 0);end--;}
}int main()
{int a[] = { 2, 3, 5, 7, 4, 6, 8 };HeapSort(a, sizeof(a) / sizeof(int));return 0;
}
二 升序 (建大堆)
//升序 建大堆
void AdjustUp(int* a, int child)
{int parent = (child - 1) / 2;while (child > 0){if (a[child] > a[parent]){Swap(&a[child], &a[parent]);child = parent;parent = (child - 1) / 2;}else{break;}}
}void AdjustDown(int* a, int n, int parent)
{int child = parent * 2 + 1;while (child < n){if (child + 1 < n && a[child + 1] > a[child]){child++;}if (a[child] > a[parent]){Swap(&a[child], &a[parent]);parent = child;child = parent * 2 + 1;}else{break;}}
}void HeapSort(int* a, int n)
{int end = n - 1;int i = 0;//建堆for (i = 1; i < n; i++){AdjustUp(a, i);}while (end > 0){Swap(&a[0], &a[end]);AdjustDown(a, end, 0);end--;}
}int main()
{int a[] = { 2, 3, 5, 7, 4, 6, 8 };HeapSort(a, sizeof(a) / sizeof(int));return 0;
}
三 优化(以升序为例)
可以用向下建堆的方法
void AdjustDown(int* a, int n, int parent)
{int child = parent * 2 + 1;while (child < n){if (child + 1 < n && a[child + 1] > a[child]){child++;}if (a[child] > a[parent]){Swap(&a[child], &a[parent]);parent = child;child = parent * 2 + 1;}else{break;}}
}//void HeapSort(int* a, int n)
//{
// int end = n - 1;
// int i = 0;
// //建堆
// for (i = 0; i < n; i++)
// {
// AdjustUp(a, i);
// }
//
// while (end > 0)
// {
// Swap(&a[0], &a[end]);
// AdjustDown(a, end, 0);
// end--;
// }
//}void HeapSort(int* a, int n)
{int end = n - 1;int i = 0;//建大堆for (i = (n-1-1) / 2; i >= 0; i--){AdjustDown(a, n, i);}while (end > 0){Swap(&a[0], &a[end]);AdjustDown(a, end, 0);end--;}
}
int main()
{int a[] = { 2, 3, 5, 7, 4, 6, 8, 65, 100, 70, 32, 50, 60};HeapSort(a, sizeof(a) / sizeof(int));return 0;
}
这样建堆的方式对时间复杂度有什么优化吗?
四 TOP-K问题
TOP - K问题：即求数据结合中前K个最大的元素或者最小的元素，一般情况下数据量都比较大。
比如：专业前10名、世界500强、富豪榜、游戏中前100的活跃玩家等。
对于Top - K问题，能想到的最简单直接的方式就是排序，但是：如果数据量非常大，排序就不太可取了(可能数据都不能一下子全部加载到内存中)。最佳的方式就是用堆来解决，基本思路如下：1. 用数据集合中前K个元素来建堆
前k个最大的元素，则建小堆
前k个最小的元素，则建大堆
2. 用剩余的N - K个元素依次与堆顶元素来比较，不满足则替换堆顶元素
将剩余N-K个元素依次与堆顶元素比完之后，堆中剩余的K个元素就是所求的前K个最小或者最大的元素。
#include<stdio.h>
#include<stdlib.h>
#include<time.h>typedef int HPDataType;void Swap(HPDataType* p1, HPDataType* p2)
{HPDataType tmp = *p1;*p1 = *p2;*p2 = tmp;
}void AdjustDown(HPDataType* a, int n, int parent)
{int child = parent * 2 + 1;while (child < n){if (child + 1 < n && a[child + 1] < a[child]){child++;}if (a[child] < a[parent]){Swap(&a[child], &a[parent]);parent = child;child = parent * 2 + 1;}else{break;}}
}void PrintTopK(const char* filename, int k)
{//1 建堆--用a中前K个元素建堆(小堆)FILE* fout = fopen(filename, "r");if (fout == NULL){perror("fopen fail");return;}int* minheap = (int*)malloc(sizeof(int) * k);if (minheap == NULL){perror("malloc fail");return;}for (int i = 0; i < k; i++){fscanf(fout, "%d", &minheap[i]);}//前K个建小堆for (int i = (k-2) / 2; i >= 0; i--){AdjustDown(minheap, k, i);}//2 将剩余n-k元素与堆顶元素比较, 满足就交换int x = 0;while (fscanf(fout, "%d", &x) != EOF){if (x > minheap[0]){//替换进堆minheap[0] = x;AdjustDown(minheap, k, 0);}}for (int i = 0; i < k; i++){printf("%d ", minheap[i]);}printf("\n");fclose(fout);}void CreateDate()
{//造数据int n = 100000;srand(time(0));const char* file = "data.txt";FILE* fin = fopen(file, "w");if (fin == NULL){perror("fopen error");return;}for (int i = 0; i < n; i++){int x = (rand() + i) % n;fprintf(fin, "%d\n", x);}fclose(fin);
}int main()
{CreateDate();PrintTopK("data.txt", 5);return 0;
}
本节对前面的二叉树基础很高, 没有理解的, 可以翻看我之前对二叉树顺序结构及其实现的章节.
继续加油!