LeetCode | Merge Intervals

题目

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].

先根据每个interval的start按升序排序，然后再像Insert Interval那样遍历一遍进行merge，不过有点小区别，在Insert Interval中，遍历到不能merge时，就可以直接移动后面的所有interval了，而这里还需要继续往后merge。

代码

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;

public class MergeIntervals {
	public ArrayList<Interval> merge(ArrayList<Interval> intervals) {
		if (intervals == null || intervals.size() == 0) {
			return intervals;
		}
		int N = intervals.size();

		// sort
		Interval[] intervalArray = intervals.toArray(new Interval[N]);
		Arrays.sort(intervalArray, new Comparator<Interval>() {
			@Override
			public int compare(Interval interval1, Interval interval2) {
				return interval1.start - interval2.start;
			}
		});

		// merge
		ArrayList<Interval> result = new ArrayList<Interval>();
		Interval mover = intervalArray[0];
		for (int i = 1; i < N; ++i) {
			if (mover.end < intervalArray[i].start) {
				result.add(mover);
				mover = intervalArray[i];
			} else {
				mover.end = Math.max(mover.end, intervalArray[i].end);
			}
		}
		result.add(mover);
		return result;
	}
}

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LeetCode | Merge Intervals

原文：http://blog.csdn.net/perfect8886/article/details/21783299