Given a collection of intervals, merge all overlapping intervals. - Java Algorithm

Java examples for Algorithm:Array

  1. HOME
  2. Java
  3. Algorithm
  4. Array

Introduction

Given a collection of intervals, merge all overlapping intervals.

For example:

Given [1,3],[2,6],[8,10],[15,18],

return [1,6],[8,10],[15,18].

Make sure the returned intervals are sorted.

Demo Code

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class Main {

  public static void main(String[] args) {
    List<Interval> intervals;

    intervals = new ArrayList<>();
    intervals.add(new Interval(1, 3));
    intervals.add(new Interval(2, 5));
    intervals.add(new Interval(1, 10));
    printRslt(new Solution().merge(intervals));

    intervals = new ArrayList<>();
    intervals.add(new Interval(1, 3));
    intervals.add(new Interval(2, 6));
    intervals.add(new Interval(8, 10));
    intervals.add(new Interval(15, 18));
    printRslt(new Solution().merge(intervals));
  }//from www. j a  v a  2  s  .co m

  private static void printRslt(List<Interval> intervals) {
    for (Interval interval : intervals) {
      System.out.print("[" + interval.start + ", " + interval.end + "], ");
    }
    System.out.println();
  }
}

class Solution {

  public List<Interval> merge(List<Interval> intervals) {
    List<Interval> rslt = new ArrayList<>();

    Collections.sort(intervals, (Interval o1, Interval o2) -> {
      return Integer.signum(o1.start - o2.start) != 0 ? Integer.signum(o1.start - o2.start)
          : Integer.signum(o1.end - o2.end);
    });

    for (int i = 0; i < intervals.size(); ++i) {
      Interval mergeInterval = intervals.get(i);

      ++i;
      while (i < intervals.size() && overlap(mergeInterval, intervals.get(i))) {
        mergeInterval.start = Math.min(mergeInterval.start, intervals.get(i).start);
        mergeInterval.end = Math.max(mergeInterval.end, intervals.get(i).end);
        ++i;
      }
      --i;

      rslt.add(mergeInterval);
    }

    return rslt;
  }

  private boolean overlap(Interval o1, Interval o2) {
    return !(Math.max(o1.start, o2.start) > Math.min(o1.end, o2.end));
  }
}

class Interval {
  int start;
  int end;

  Interval() {
    start = end = 0;
  }

  Interval(int s, int e) {
    start = s;
    end = e;
  }
}

Related Tutorials