最小的k个数（无序）
最小的K个数
http://www.nowcoder.com/questionTerminal/6a296eb82cf844ca8539b57c23e6e9bf
import java.util.ArrayList;
public class Solution {
public ArrayList<Integer> GetLeastNumbers_Solution(int [] input, int k) {
ArrayList<Integer> list = new ArrayList<Integer>();
int[] res = new int[k];
int temp;
while (input == null || k<=0 || k > input.length) {
return list;
}
for (int i = k / 2 - 1; i >= 0; i--) {
adjustHeap(input, i, k-1);
}
for (int i = k; i < input.length; i++) {
if (input[i] < input[0]) {
input[0] = input[i];
adjustHeap(input, 0, k-1);
}
}
for (int i = 0;i < k; i++) {
list.add(input[i]);
}
return list;
}
public void adjustHeap(int []arr, int start, int end) {
int temp = arr[start];
for (int child = start * 2 + 1; child <= end; child = child * 2 + 1) {
if (child + 1 <= end && arr[child]<arr[child + 1]) {
child++;
}
if (arr[child] < temp) {
break;
}
arr[start] = arr[child];
start = child;
}
arr[start] = temp;
}
} 思路是：构造节点数为k的大顶堆维护数组最小的k个数，只要有数比堆内最大值（根节点）还小，就将根节点换出，同时调整大顶堆
若要有序的k个数，则需要将整个数组构建小顶堆，而后取出k个根节点组成最小k个数
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