海量数据处理之Top K问题

向右看齐 2021-11-29 10:08 297阅读 0赞

#### 海量数据处理之Top K问题 ####

#### 1、问题描述 ####

将班里所有同学喜欢吃的水果存储于一个数组中。  
要求：统计出所有水果出现的次数，并且求出大家最喜欢吃的前k种水果。

#### 2、Main.c ####

#include "excel.h"
    #include <windows.h>
    
    int main()
    {
        
    	//Generate();
    	FindTopK();
    	system("pause");
    
    	return 0;
    }

#### 3、excel.h ####

#define _CRT_SECURE_NO_WARNINGS
    #pragma once
    
    #include <stdlib.h>
    #include <stdio.h>
    #include <string.h>
    #include <assert.h>
    
    static char * fruits[] = {
        
    	"apple", "pear", "banana", "orange"
    };
    
    int size = sizeof(fruits) / sizeof(char *);
    
    void Generate()
    {
        
    	srand(20190711);
    	for (int i = 0; i < 60; i++) {
        
    		printf("    \"%s\", \n", fruits[rand() % size]);
    	}
    }
    
    static char * favor_fruits[] = {
        
    	"pear",
    	"pear",
    	"apple",
    	"banana",
    	"apple",
    	"apple",
    	"pear",
    	"banana",
    	"banana",
    	"apple",
    	"banana",
    	"banana",
    	"pear",
    	"orange",
    	"banana",
    	"apple",
    	"apple",
    	"orange",
    	"apple",
    	"orange",
    	"orange",
    	"banana",
    	"pear",
    	"banana",
    	"pear",
    	"pear",
    	"apple",
    	"apple",
    	"orange",
    	"orange",
    	"pear",
    	"banana",
    	"apple",
    	"orange",
    	"banana",
    	"banana",
    	"orange",
    	"pear",
    	"orange",
    	"pear",
    	"pear",
    	"apple",
    	"apple",
    	"orange",
    	"apple",
    	"banana",
    	"orange",
    	"pear",
    	"banana",
    	"apple",
    	"apple",
    	"apple",
    	"banana",
    	"banana",
    	"pear",
    	"pear",
    	"apple",
    	"pear",
    	"banana",
    	"apple"
    };
    
    int favor_size = sizeof(favor_fruits) / sizeof(char *);
    
    struct ResultItem {
        
    	char fruit[20];	// key
    	int count;		// value
    };
    
    
    struct Node {
        
    	struct ResultItem data;
    	struct Node *left;
    	struct Node *right;
    };
    
    void Insert(struct Node **root, char *fruit)
    {
        
    	struct Node *node = (struct Node *)malloc(sizeof(struct Node));
    	strcpy(node->data.fruit, fruit);
    	node->data.count = 1;
    	node->left = node->right = NULL;
    
    	if (*root == NULL) {
        
    		*root = node;
    		return;
    	}
    
    	int r = strcmp((*root)->data.fruit, fruit);
    	assert(r != 0);	// 外面保证一定不会插入 fruit 重复的值
    
    	if (r < 0) {
        
    		Insert(&(*root)->left, fruit);
    	}
    	else {
        
    		Insert(&(*root)->right, fruit);
    	}
    }
    
    
    struct ResultItem * Find(struct Node *root, char *fruit)
    {
        
    	if (root == NULL) {
        
    		return NULL;
    	}
    
    	int r = strcmp(root->data.fruit, fruit);
    	if (r == 0) {
        
    		return &root->data;	// ?
    	}
    
    	if (r < 0) {
        
    		return Find(root->left, fruit);
    	}
    	else {
        
    		return Find(root->right, fruit);
    	}
    }
    
    void PreOrder(struct ResultItem * result, struct Node *root)
    {
        
    	if (root == NULL) {
         return; }
    
    	int i;
    	switch (root->data.fruit[0]) {
        
    	case 'a':
    		// apple
    		i = 0;
    		break;
    	case 'b':
    		// banana
    		i = 1;
    		break;
    	case 'o':
    		// orange
    		i = 2;
    		break;
    	case 'p':
    		// pear
    		i = 3;
    		break;
    	}
    
    	result[i] = root->data;
    	printf("%s %d\n", root->data.fruit, root->data.count);
    	
    	PreOrder(result, root->left);
    	PreOrder(result, root->right);
    }
    
    struct ResultItem * StaticsFruitCount(char *fruits[], int size, int *pResultSize)
    {
        
    	// 假设有个集合 Set
    	// 1. 顺序表
    	// 2. 二叉搜索树
    	// 3. 哈希表
    	// 正常考虑，这个数据量，有顺序表更合适
    	// 此处强行用二叉搜索树
    	struct Node *root = NULL;
    
    	for (int i = 0; i < size; i++) {
        
    		char *fruit = fruits[i];
    
    		struct ResultItem *item = Find(root, fruit);
    		if (item != NULL) {
        
    			// 如果已经有过，出现次数 + 1
    			item->count++;
    		}
    		else {
        
    			// 如果没有出现过，放进去，次数是 1 次
    			Insert(&root, fruit);
    		}
    	}
    
    	// 前序遍历的方式取出来
    	struct ResultItem *result = (struct ResultItem *)malloc(4 * sizeof(struct ResultItem));
    	PreOrder(result, root);
    
    	*pResultSize = 4;
    	return result;
    }
    
    void ArrayAdjustDown(struct ResultItem array[], int size, int root)
    {
        
    	int parent = root;
    
    	while (1) {
        
    		// 先判断有没有孩子（叶子结点）
    		// 数组角度去想	-> 孩子的下标是否越界
    		// 只要判断左孩子的下标（因为是完全二叉树)
    		int left = parent * 2 + 1;
    		if (left >= size) {
        
    			// 越界，没有左孩子，也肯定没有右孩子
    			return;
    		}
    
    		// 一定有左孩子
    		int maxChild = left;
    		if (2 * parent + 2 < size && array[2 * parent + 2].count < array[left].count) {
        
    			// 前一个条件是判断右孩子有没有
    			// 后一个条件判读是右孩子是否比左孩子大
    			maxChild = 2 * parent + 2;
    		}
    
    		if (array[parent].count < array[maxChild].count) {
        
    			return;
    		}
    
    		// 交换 root 和 maxChild 下标所在的值
    		struct ResultItem t = array[parent];
    		array[parent] = array[maxChild];
    		array[maxChild] = t;
    
    		parent = maxChild;
    	}
    }
    
    struct ResultItem * TopK(struct ResultItem *result, int size, int k)
    {
        
    	struct ResultItem *heap = (struct ResultItem *)malloc(sizeof(struct ResultItem) * k);
    	assert(heap);
    
    	for (int i = 0; i < k; i++) {
        
    		heap[i] = result[i];
    	}
    
    	for (int j = (k - 2) / 2; j >= 0; j--) {
        
    		ArrayAdjustDown(heap, k, j);
    	}
    
    	for (int m = k; m < size; m++) {
        
    		if (result[m].count > heap[0].count) {
        
    			heap[0] = result[m];
    			ArrayAdjustDown(heap, k, 0);
    		}
    	}
    
    	return heap;
    }
    
    void FindTopK()
    {
        
    	int result_size;
    
    	// result 有多少个元素, result_size 个
    	struct ResultItem *result = StaticsFruitCount(favor_fruits, favor_size, &result_size);
    
    	// topk 有多少个元素，答案就是 k 个
    	struct ResultItem *topk = TopK(result, result_size, 2);
    
    	for (int i = 0; i < 2; i++) {
        
    		printf("%s %d\n", topk[i].fruit, topk[i].count);
    	}
    }