假设你正在给一组N个区间（重psented为左，右坐标$ P $）和M点。对于每一个点P算法应该发现间隔的数目，以该P所属于

Suppose you are given a set of N intervals (represented as left and right coordinates) and M points. For each point P algorithm should find number of intervals to which P belongs.

下面是我的算法：

1）把左，间隔左和右的数组分别右侧坐标

1) Put left and right coordinates of intervals in "left" and "right" arrays respectively

2）排序左，交换条目正确的同步

2) Sort "left", exchanging entries simultaneously with "right"

3）给定的点P，找到的最大i使得留下[1] - ; = P

3) Given a point P, find a maximum i such that left[i] <= P

4）对于每一个J＆LT; = I，加1，如果结果正确的研究[J]> = P

4) For each j <= i, add 1 to result if right[j] >= P

5）返回结果

实现在Java中：

import java.util.*;

class Intervals {

    public  static int count(int[] left, int[] right, int point) {
        int k = find(left, point), result = 0;
        for (int i=0; i < k; i++)
            if (point <= right[i]) result++;
        return result;
    }


    private static int find(int[] a, int point) {
        if (point < a[0]) return -1;
        int i = 0;
        while (i < a.length && a[i] <= point) i++;
        return i;
    }

    private static void sort(int[] a, int[] b) {
        sort(a, b, 0, a.length-1);
    }

    private static void sort(int[] left, int[] right, int lo, int hi) {
        if (hi <= lo) return;
        int lt = lo, gt = hi;
        exchange(left, right, lo, lo + (int) (Math.random() * (hi-lo+1)));
        int v = left[lo];
        int i = lo;
        while (i <= gt) {
            if      (left[i] < v) exchange(left, right, lt++, i++);
            else if (left[i] > v) exchange(left, right, i, gt--);
            else                  i++;
        }
        sort(left, right, lo, lt-1);
        sort(left, right, gt+1, hi);
    }

    private static void exchange(int[] left, int[] right, int i, int j) {
        int temp  = left[i];
        left[i]   = left[j];
        left[j]   = temp;
        temp      = right[i];
        right[i]  = right[j];
        right[j]  = temp;
    }

    private static boolean less(int[] a, int i, int j) {
        return a[i] < a[j];
    }


    public static void main(String[] args) {
        int n       = Integer.parseInt(args[0]);
        int m       = Integer.parseInt(args[1]);
        int[] left  = new int[n];
        int[] right = new int[n];
        Random r    = new Random();
        int MAX     = 100000;
        for (int i = 0; i < n; i++) {
            left[i] = r.nextInt(MAX);
            right[i] = left[i] + r.nextInt(MAX/4);
        }
        sort(left, right);
        for (int i=0; i < m; i++)
            System.out.println(count(left, right, r.nextInt(MAX)));
    }
}

这code有没有通过一些测试，而我试图找到一个错误。问题是，我居然不知道什么是输入数据在这些测试中使用。

This code has not passed some test, and I'm trying to find a bug. Point is that I actually don't know what input data used in these tests.

感谢。

推荐答案

也许不是你要找的，但可能是回答过这个问题，有人来了新的一天。答案

Probably not the answer you are looking for but may be the answer to someone coming across this question another day.

如果您计划查询一个相当静态的范围集合经常那么你不妨考虑一个区间树。

If you are planning to query a fairly static set of ranges often then you may wish to consider an Interval Tree.

public class IntervalTree<T extends IntervalTree.Interval> {
  // My intervals.

  private final List<T> intervals;
  // My center value. All my intervals contain this center.
  private final long center;
  // My interval range.
  private final long lBound;
  private final long uBound;
  // My left tree. All intervals that end below my center.
  private final IntervalTree<T> left;
  // My right tree. All intervals that start above my center.
  private final IntervalTree<T> right;

  public IntervalTree(List<T> intervals) {
    if (intervals == null) {
      throw new NullPointerException();
    }

    // Initially, my root contains all intervals.
    this.intervals = intervals;

    // Find my center.
    center = findCenter();

    /*
     * Builds lefts out of all intervals that end below my center.
     * Builds rights out of all intervals that start above my center.
     * What remains contains all the intervals that contain my center.
     */

    // Lefts contains all intervals that end below my center point.
    final List<T> lefts = new ArrayList<T>();
    // Rights contains all intervals that start above my center point.
    final List<T> rights = new ArrayList<T>();

    long uB = Long.MIN_VALUE;
    long lB = Long.MAX_VALUE;
    for (T i : intervals) {
      long start = i.getStart();
      long end = i.getEnd();
      if (end < center) {
        lefts.add(i);
      } else if (start > center) {
        rights.add(i);
      } else {
        // One of mine.
        lB = Math.min(lB, start);
        uB = Math.max(uB, end);
      }
    }

    // Remove all those not mine.
    intervals.removeAll(lefts);
    intervals.removeAll(rights);
    uBound = uB;
    lBound = lB;

    // Build the subtrees.
    left = lefts.size() > 0 ? new IntervalTree<T>(lefts) : null;
    right = rights.size() > 0 ? new IntervalTree<T>(rights) : null;

    // Build my ascending and descending arrays.
    /**
     * @todo Build my ascending and descending arrays.
     */
  }

  /*
   * Returns a list of all intervals containing the point.
   */
  List<T> query(long point) {
    // Check my range.
    if (point >= lBound) {
      if (point <= uBound) {
        // In my range but remember, there may also be contributors from left or right.
        List<T> found = new ArrayList<T>();
        // Gather all intersecting ones.
        // Could be made faster (perhaps) by holding two sorted lists by start and end.
        for (T i : intervals) {
          if (i.getStart() <= point && point <= i.getEnd()) {
            found.add(i);
          }
        }

        // Gather others.
        if (point < center && left != null) {
          found.addAll(left.query(point));
        }
        if (point > center && right != null) {
          found.addAll(right.query(point));
        }

        return found;
      } else {
        // To right.
        return right != null ? right.query(point) : Collections.<T>emptyList();
      }
    } else {
      // To left.
      return left != null ? left.query(point) : Collections.<T>emptyList();
    }

  }

  private long findCenter() {
    //return average();
    return median();
  }

  /**
   * @deprecated Causes obscure issues.
   * @return long
   */
  @Deprecated
  protected long average() {
    // Can leave strange (empty) nodes because the average could be in a gap but much quicker.
    // Don't use.
    long value = 0;
    for (T i : intervals) {
      value += i.getStart();
      value += i.getEnd();
    }
    return intervals.size() > 0 ? value / (intervals.size() * 2) : 0;
  }

  protected long median() {
    // Choose the median of all centers. Could choose just ends etc or anything.
    long[] points = new long[intervals.size()];
    int x = 0;
    for (T i : intervals) {
      // Take the mid point.
      points[x++] = (i.getStart() + i.getEnd()) / 2;
    }
    Arrays.sort(points);
    return points[points.length / 2];
  }

  void dump() {
    dump(0);
  }

  private void dump(int level) {
    LogFile log = LogFile.getLog();
    if (left != null) {
      left.dump(level + 1);
    }
    String indent = "|" + StringUtils.spaces(level);
    log.finer(indent + "Bounds:- {" + lBound + "," + uBound + "}");
    for (int i = 0; i < intervals.size(); i++) {
      log.finer(indent + "- " + intervals.get(i));
    }
    if (right != null) {
      right.dump(level + 1);
    }

  }

  /*
   * What an interval looks like.
   */
  public interface Interval {

    public long getStart();

    public long getEnd();
  }

  /*
   * A simple implemementation of an interval.
   */
  public static class SimpleInterval implements Interval {

    private final long start;
    private final long end;

    public SimpleInterval(long start, long end) {
      this.start = start;
      this.end = end;
    }

    public long getStart() {
      return start;
    }

    public long getEnd() {
      return end;
    }

    @Override
    public String toString() {
      return "{" + start + "," + end + "}";
    }
  }

  /**
   * Not called by App, so you will have to call this directly.
   * 
   * @param args 
   */
  public static void main(String[] args) {
    /**
     * @todo Needs MUCH more rigorous testing.
     */
    // Test data.
    long[][] data = {
      {1, 2},
      {2, 9},
      {4, 8},
      {3, 5},
      {7, 9},};
    List<Interval> intervals = new ArrayList<Interval>();
    for (long[] pair : data) {
      intervals.add(new SimpleInterval(pair[0], pair[1]));
    }
    // Build it.
    IntervalTree<Interval> test = new IntervalTree<Interval>(intervals);

    // Test it.
    System.out.println("Normal test: ---");
    for (long i = 0; i < 10; i++) {
      List<Interval> intersects = test.query(i);
      System.out.println("Point " + i + " intersects:");
      for (Interval t : intersects) {
        System.out.println(t.toString());
      }
    }

    // Check for empty list.
    intervals.clear();
    test = new IntervalTree<Interval>(intervals);
    // Test it.
    System.out.println("Empty test: ---");
    for (long i = 0; i < 10; i++) {
      List<Interval> intersects = test.query(i);
      System.out.println("Point " + i + " intersects:");
      for (Interval t : intersects) {
        System.out.println(t.toString());
      }
    }

  }
}