LeetCode OJ 56. Merge Intervals 贪心法求解

题目链接：https://leetcode.com/problems/merge-intervals/

56. Merge Intervals

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Question

Total Accepted: 60386 Total
Submissions: 244495 Difficulty: Hard

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]

.

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给出一个区间数组，要求将重叠的区间合并。是个hard级别的题目，但个人觉得很easy。

显然greedy求解。将所有的区间按start排序，start相同则以end排序。然后从左向右逐个合并。注意一个边界条件，即，左边的大区间包含了右边的小区间。

我的AC代码

public class MergeIntervals {

	public List<Interval> merge(List<Interval> intervals) {
		if (intervals == null || intervals.size() <= 1) {
			return intervals;
		}
		Collections.sort(intervals, new Comparator<Interval>() {

			@Override
			public int compare(Interval o1, Interval o2) {
				int r = o1.start - o2.start;
				if (r != 0)
					return r;
				return o1.end - o2.end;
			}
		});
		List<Interval> r = new ArrayList<Interval>();
		Interval in = intervals.get(0);
		for (int i = 1; i < intervals.size(); i++) {
			Interval cur = intervals.get(i);
			if (cur.start <= in.end && in.end <= cur.end) {
				in.end = cur.end;
			} else if(cur.start > in.end){
				r.add(in);
				in = cur;
			} else if(cur.start <= in.end && in.end >= cur.end){
				continue;
			}
		}
		r.add(in);
		return r;
	}
}