题目描述
输入n个整数,找出其中最小的K个数。例如输入4,5,1,6,2,7,3,8这8个数字,则最小的4个数字是1,2,3,4,。
解题思路
用最大堆保存这k个数,每次只和堆顶比,如果比堆顶小,删除堆顶,新数入堆。
import java.util.ArrayList;
import java.util.Comparator;
import java.util.PriorityQueue;
public class $40_GetLeastNumbers_Solution {
public static void main(String[] args) {
$40_GetLeastNumbers_Solution solution = new $40_GetLeastNumbers_Solution();
int[] arr = {4, 5, 1, 6, 2, 7, 3, 8};
int k = 4;
ArrayList<Integer> res = solution.GetLeastNumbers_Solution(arr, k);
System.out.println("res:" + res); // [4,3,1,2]
}
public ArrayList<Integer> GetLeastNumbers_Solution(int[] input, int k) {
ArrayList<Integer> result = new ArrayList<Integer>();
int length = input.length;
if (k > length || k == 0) {
return result;
}
PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(k, new Comparator<Integer>() {
@Override
public int compare(Integer o1, Integer o2) {
return o2.compareTo(o1); // 并非从小到大排序,确保维持队首值为最大值
}
});
for (int i = 0; i < length; i++) {
// 如果队列中元素个数未到k,直接添加
if (maxHeap.size() != k) {
maxHeap.offer(input[i]);
}
// 达到k数,判断下一个值是否小于队首值,小于则将队首值删除,将此值添加
else if (input[i] < maxHeap.peek()) {
Integer temp = maxHeap.poll();
temp = null; //GC回收
maxHeap.offer(input[i]);
}
}
for (Integer integer : maxHeap) {
result.add(integer);
}
return result;
}
}