Java去除一个Map中value值最小的K个记录

import java.util.*;public class TestHeapSorted {    public static void main(String[] args) {        Map<String, String> testMap = new HashMap<String, String>();        testMap.put("tag2", "2");        testMap.put("tag1", "1");        testMap.put("tag3", "3");        testMap.put("tag5", "5");        testMap.put("tag4", "4");        testMap.put("tag7", "7");        testMap.put("tag9", "9");        testMap.put("tag8", "8");        testMap.put("tag6", "6");        Object[] testObjectValue = testMap.values().toArray();        TestHeapSorted testHeapSorted = new TestHeapSorted();        System.out.println(testHeapSorted.GetLeastNumbers_Solution(testObjectValue, (3 + 1)));        List<Double> testList = testHeapSorted.GetLeastNumbers_Solution(testObjectValue, (3 + 1));        Iterator<Map.Entry<String, String>> it = testMap.entrySet().iterator();        while (it.hasNext()) {            Map.Entry<String, String> entry = it.next();            for (int i = 0; i < testList.size(); i++) {                if (Double.parseDouble(entry.getValue()) == testList.get(i)) {                    testList.remove(i);                    it.remove();                    break;                }            }        }        for (String key : testMap.keySet()) {            System.out.println(key + "=>" + testMap.get(key));        }        System.out.println(testList.size());    }    public ArrayList<Double> GetLeastNumbers_Solution(Object[] input, int k) {        ArrayList<Double> res = new ArrayList<Double>();        if (input == null || input.length == 0 || input.length < k) {            return res;        }        if (k == 0)            return res;        double[] maxHeap = new double[k];        //初始化堆        for (int i = 0; i < maxHeap.length; i++) {            maxHeap[i] = Double.parseDouble(input[i].toString());        }        //将初始化的堆调整为最大堆。最大堆：根节点的值总是大于子树中任意节点的值        // 注意是--i，从下往上        for (int i = (maxHeap.length - 1) / 2; i >= 0; i--) {            adjustHeap(maxHeap, i);        }        //遍历初始数组不断调整最大堆        for (int i = k; i < input.length; i++) {            if (maxHeap[0] > Double.parseDouble(input[i].toString())) {                maxHeap[0] = Double.parseDouble(input[i].toString());                adjustHeap(maxHeap, 0);   // 是0位置，即根节点的值为最大值            }        }        for (int i = 0; i < maxHeap.length; i++) {            res.add(maxHeap[i]);        }        return res;    }    static void adjustHeap(double maxHeap[], int i) {        int index = i;        int lchild = 2 * i + 1;  //i的左孩子节点序号        int rchild = 2 * i + 2; //i的右孩子节点序号        if (index <= (maxHeap.length - 1) / 2) {            //寻找子节点中最大的节点            if (lchild < maxHeap.length && maxHeap[index] < maxHeap[lchild]) { //左孩子大于index                index = lchild;            }            if (rchild < maxHeap.length && maxHeap[index] < maxHeap[rchild]) { //右孩子大于index                index = rchild; //此时index为左右孩子中较大的那个            }            if (i != index) {                //将节点与最大的子节点交换                double tmp = maxHeap[index];                maxHeap[index] = maxHeap[i];                maxHeap[i] = tmp;                //交换后，子树可能不满足最大推，递归调整。                adjustHeap(maxHeap, index); //因为交换下来一个小的数字，要调整的是被交换的子树            }        }    }}

答案：

[4.0, 3.0, 2.0, 1.0]
tag5=>5
tag8=>8
tag9=>9
tag6=>6
tag7=>7
0