(微软面试100题)查找最小的K个元素

问题描述：

输入n个整数，输出其中最小的k个。
例如输入1，2，3，4，5，6，7和8这8个数字，
则最小的4个数字为1，2，3和4。

解题思路：

构建一个K个元素的堆，最终的算法复杂度为K*log(n)

#include "stdafx.h"
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>


/** 
 * @brief  max_heapify, filter up a heap
 * 
 * @param heap[]
 * @param s         start index
 * @param e         end index
 * 
 * @return   0
 */
int max_heapify(int heap[], int s, int e)
{
	int rc;
	int i;

	rc = heap[s];
	for (i = 2*s + 1; i <= e; i = 2*i + 1)
	{
		if ((i + 1) <= e && heap[i] <= heap[i+1]){
			i++;
		}
		if (heap[i] < rc){
			break;
		}
		heap[s] = heap[i];
		s = i;
	}
	heap[s] = rc;
	return 0;
}

/*
 * Build a max heap
 */
int build_max_heap(int heap[], int heap_size)
{
	int i;
	for (i = heap_size/2-1; i >= 0; i--)
	{
		max_heapify(heap, i, heap_size-1);
	}
	return 0;
}

/*
 *  Add one value to heap and adjust it
 */
int add_one_vaule(int heap[], int heap_size, int val)
{
    /* We need k-min values */
	if (val < heap[0]){
		heap[0] = val;
		max_heapify(heap, 0, heap_size-1);
	}
	return 0;
}

/*
 *  Find K min values
 */
int find_k_min(int data[], int len, int k)
{
	int *heap;
	int i;
	heap = (int *)malloc(k * sizeof(int));
	if (heap == NULL){
		printf("Malloc error!\n");
		return -1;
	}
	for (i = 0; i < k; i++)
	{
		heap[i] = data[i];
	}
	build_max_heap(heap, k);
	for (; i < len; i++)
	{
		add_one_vaule(heap, k, data[i]);
	}
	for (i = 0; i < k; i++)
	{
		printf("%d,", heap[i]);
	}
	printf("\n");
	free(heap);
	return 0;
}

int main()
{
	int data[] = {8, 3, 2, 10, 5, 7, 9, 1, 4, 0, 7};
	int len = sizeof(data) / sizeof(data[0]);
	find_k_min(data, len, 4);
	return 0;
}