[leetcode] Merge Intervals

本文介绍如何将一组区间合并为最少数量的不相交区间。

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


/**
 * 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].
 * 
 * 
 * @author Chris
 *
 */


class Interval  implements Comparable<Interval>{
private int start;
private int end;
public Interval() {

}

public Interval(int start, int end) {
this.start = start;
this.end = end;
}



public void setStart(int start) {
this.start = start;
}


public void setEnd(int end) {
this.end = end;
}


public int getStart() {
return start;
}


public int getEnd() {
return end;
}


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


@Override
public int compareTo(Interval o) {
return this.getStart() - o.getStart();
}

}


public class MergeIntervals {

static List<Interval> getMergeIntervals(List<Interval> intervals) {
if (intervals == null || intervals.size() == 0) {
return null;
}
List<Interval> result = new ArrayList<Interval>();

Collections.sort(intervals);
System.out.println(intervals);

Interval toAdd = intervals.get(0);
for (int i = 1; i < intervals.size(); i++) {
if(intervals.get(i).getStart() > toAdd.getEnd()) {
result.add(toAdd);
toAdd = intervals.get(i);
} else {
toAdd.setStart(toAdd.getStart());
int oldEnd = toAdd.getEnd();
toAdd.setEnd(Math.max(intervals.get(i).getEnd(), oldEnd));
}
}

if (toAdd != null) {
result.add(toAdd);
}

return result;
}


/**
* @param args
*/
public static void main(String[] args) {
List<Interval> intervals = new ArrayList<Interval>();
intervals.add(new Interval(1,3));
intervals.add(new Interval(8,10));
intervals.add(new Interval(15,18));
intervals.add(new Interval(2,6));

System.out.println("result = " + getMergeIntervals(intervals));


}


}

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