algorithm - 范围分割问题

Question

伙计们，

很久以前就听说过这个问题。想到发布它，以获得一些关于使用某种构造或其他有效手段（可能是专门的树）这样做的看法

给定一组成对的范围 (5,18) (12,23) (15,30)

将它们分成所有可能的子范围，这些子范围与集合中的其他范围重叠。像 (5,11) (12,14) (15,18) (19,23) (24,30)

谢谢大家，谢谢...

拉詹...

PS这是一个标准问题，如果是，想知道它的名字

Answer 1

将所有范围端点放入列表中，但将它们标记为起点/终点。

[(5, S), (18, E), (12, S), (23, E), (15, S), (30, E)]

按位置对它们进行排序，通过将起点放在终点之前来打破平局。

[(5, S), (12, S), (15, S), (18, E), (23, E), (30, E)]

然后你可以通过遍历这个列表来计算出范围，跟踪到目前为止我们已经处理了多少个起点-终点。如果我们看到一个起点，那就是一个新的输出范围的起点。如果我们的计数是正数，我们必须先结束当前范围。如果我们看到一个终点，则结束当前范围。

Answer 2

也许我遗漏了一些东西，但这似乎是一个简单的解决方案：将所有数字放入 C++ STL 集容器中。它们将按升序自动排序。所以他们会以这种方式读出：5,12,15,18,23,30 所以如果你能容忍重叠，那么：(5,12) (12,15) (15,18), (18,23) ( 23,30) 是通过将每个数字重复两次（除了第一个和最后一个）然后将两个数字分组来构造的范围。

如果您不能容忍重叠，则可以通过在列表中放置一个递增的数字而不是重复它来构建范围。

Answer 3

好的，经过一番修改后，我能够实现一个明显有效的版本。因此，对于那些正在寻找可行解决方案的人来说，这是我的：

private static class RangeVal implements Comparable<RangeVal> {
    public final BigInteger value;
    public int count;

    public RangeVal(BigInteger value, int count) {
        super();
        this.value = value;
        this.count = count;
    }

    @Override
    public String toString() {
        return value + (isStart() ? "S" : "E") + count;
    }

    @Override
    public int compareTo(RangeVal o) {
        // Sort by value first
        int v = value.compareTo(o.value);
        if (v != 0)
            return v;
        // Then sort Starts before ends
        return -count;
    }

    public boolean isStart() {
        return count > 0;
    }

}

/**
 * Sort a List of ranges by their number, then start/end and merge multiple
 * start/ends
 * 
 * @param temp
 *            a list of RangeVal which can be unsorted
 */
private static void preSort(List<RangeVal> temp) {
    Collections.sort(temp);
    RangeVal last = null;
    for (Iterator<RangeVal> iterator = temp.iterator(); iterator.hasNext();) {
        RangeVal rangeVal = iterator.next();
        if ((last != null) && last.value.equals(rangeVal.value) && (last.isStart() == rangeVal.isStart())) {
            iterator.remove();
            last.count += rangeVal.count;
        } else
            last = rangeVal;
    }
}

/**
 * Splits a list into ValueRange Objects that do not overlap each other, but
 * fully represent the ranges given by value
 * 
 * @param value
 *            a list of RangeVal Objects that need to be split
 * @return
 */
private static SortedSet<ValueRange> split(List<RangeVal> value) {
    preSort(value);
    SortedSet<ValueRange> res = new TreeSet<ValueRange>();
    int count = 0;
    BigInteger start = null;
    for (RangeVal rangeVal : value) {
        count += rangeVal.count;
        if (rangeVal.isStart()) {
            if (start != null) {
                //If there was an unended start, then we have to end it just one before the new start
                res.add(new ValueRange(start, rangeVal.value.subtract(BigInteger.ONE)));
            }
            //Set the start to the current Element
            start = rangeVal.value;
        } else {
            //End the current range at this Element
            res.add(new ValueRange(start, rangeVal.value));
            if (count > 0) {
                //If we expect another end later, the element following this will have to start one after
                start = rangeVal.value.add(BigInteger.ONE);
            } else
                //No new range anymore
                start = null;
        }
    }
    return res;
}

public static void main(String[] args) {
    // 5->8 9->10 11
    System.out.println(split(createRanges(5, 8, 9, 10, 11, 11)));
    // 5, 6->7, 8, 9, 10
    System.out.println(split(createRanges(5, 10, 6, 8, 8, 9)));
    System.out.println(split(createRanges(5, 10, 6, 8, 8, 9, 6, 9)));
    System.out.println(split(createRanges(5, 10, 6, 8, 8, 9, 6, 9, 6, 11, 8, 9)));
    System.out.println(split(createRanges(5, 10, 6, 8, 8, 9, 6, 9, 6, 11, 8, 9, 14, 18)));
}

private static List<RangeVal> createRanges(int... r) {
    List<RangeVal> temp = new LinkedList<RangeVal>();
    for (int i = 0; i < r.length; i++) {
        temp.add(new RangeVal(BigInteger.valueOf(r[i]), (i % 2) == 0 ? 1 : -1));
    }
    System.out.println("HDLSimulator.createRanges()" + temp);
    return temp;
}