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java如何使用LinkedHashMap获取子映射?

目前,我正在使用TreeMap来存储一些x和y坐标,但是与ArrayListHashMap相比,迭代非常慢。我之所以使用它,是因为我需要subMap()方法,这样我就可以在确定的范围内获得X值,即使精确的X值(键)不存在

LinkedHashMap的速度几乎与HashMap相同,我可以按插入顺序迭代键(我需要插入顺序或比较器顺序,就像在TreeMap中那样),但我没有submap()方法。在树状图中,我可以快速生成子图

是否有任何数据结构或某种方式可以比TreeMap更快地存储有序值(通过插入顺序或比较器),即使精确值不在映射中,也可以在一定范围内获取子映射?我的意思是,也许我想要2到25之间的值,但是2不存在,最近的值是3,所以它将返回一个从3到25的子映射。或者以某种方式将此功能添加到LinkedHashMap


共 (2) 个答案

  1. # 1 楼答案

    今天我终于找到了问题的答案。经过几次测试HashMapLinkedHashMapTreeMapArrayList慢得多,我想使用它们只是为了创建subMaps()。因此,我创建了一个扩展ArrayList的新类,它给了我非常好的性能,在this答案的帮助下,我创建了一种通过值而不是索引获取子列表的快速方法。以下是完整的课程:

    /**
     * The purpose of this class is to be a faster replacement to a {@link java.util.TreeMap} with
     *  the ability to get sublist containing a range of x values. ArrayList access time is O(1) while
     *  {@link java.util.TreeMap} is O(log(n)). When large data is handled the impact on performance is
     *  noticeable.
     */
    public class XYDataset extends ArrayList<PointValue> {
    
        private final float COMPARISON_THRESHOLD = 0.01f;
    
        final Comparator<PointValue> comparator = new Comparator<PointValue>() {
            @Override
            public int compare(PointValue lhs, PointValue rhs) {
                if (Math.abs(lhs.getX() - rhs.getX()) < COMPARISON_THRESHOLD) return 0;
                return lhs.getX() < rhs.getX() ? -1 : 1;
            }
        };
    
        public XYDataset(int capacity) {
            super(capacity);
        }
    
        public XYDataset() {
        }
    
        public XYDataset(Collection<? extends PointValue> collection) {
            super(collection);
        }
    
        @Override
        public List<PointValue> subList(int start, int end) {
            return super.subList(start, end);
        }
    
        /**
         * Generate a sublist containing the range of x values passed
         * @param x1 lower x value
         * @param x2 upper x value
         * @return sublist containing x values from x1 to x2
         */
        public List<PointValue> subList(float x1, float x2){
            /**
             * Collections.binarySearch() returns the index of the search key, if it is contained in the list;
             *  otherwise it returns (-(insertion point) - 1).
             * The insertion point is defined as the point at which the key would be inserted into the list:
             *  the index of the first element greater than the key, or list.size() if all elements in the list
             *  are less than the specified key. Note that this guarantees that the return value will be >= 0 if
             *  and only if the key is found.
             */
            int n1 = Collections.binarySearch(this, new PointValue(x1, 0), comparator);
            int n2 = Collections.binarySearch(this, new PointValue(x2, 0), comparator);
    
            /**
             * Example, we assume the list is sorted. Based on (https://stackoverflow.com/questions/19198586/search-sorted-listlong-for-closest-and-less-than)
             *
             * long X = 500;
             * List<Long> foo = new Arraylist<>();
             * foo.add(450L);
             * foo.add(451L);
             * foo.add(499L);
             * foo.add(501L);
             * foo.add(550L);
             *
             * If we search for something that isn't in the list you can work backward from the return value
             *  to the index you want. If you search for 500 in your example list, the algorithm would return (-3 - 1) = -4.
             * Thus, you can add 1 to get back to the insertion point (-3), and then multiply by -1 and subtract 1 to get
             *  the index BEFORE the first element GREATER than the one you searched for, which will either be an index that
             *  meets your 2 criteria OR -1 if all elements in the list are greater than the one you searched for.
             */
            if(n1 < 0) n1 = -n1-1;
            if(n2 < 0) n2 = -n2-1;
    
            return this.subList(n1, n2);
        }
    }
    

    PointValue只是一个包含x和y坐标的类。现在我只要调用subList()传递我想要的x坐标范围。在我的例子中,插入顺序也被排序,这对于使用Collections.binarySearch()很重要

  2. # 2 楼答案

    听起来你需要一个树形图,它的迭代速度不比LinkedHashMap慢多少,并且可以实现你真正想要的。由于HashMap是无序的,所以subMap没有任何意义