Merge Intervals

于 2015-06-20 13:41:27 发布 · 62 阅读

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本文介绍了一种有效的算法来合并一系列可能存在重叠的区间。通过排序和遍历的方式，该算法可以将所有重叠的区间合并为一个区间，最终返回一个没有重叠的区间列表。例如，给定区间 [1,3],[2,6],[8,10],[15,18]，合并后的结果为 [1,6],[8,10],[15,18]。

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].

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    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) {
				// TODO Auto-generated method stub
				return o1.start - o2.start;
			}
    		
		});
    	List<Interval> res = new ArrayList<>();
    	Interval pre = intervals.get(0);
    	for (int i = 1; i < intervals.size(); i++) {
    		Interval cur = intervals.get(i);
    		if (pre.end >= cur.start) {
    			Interval merge = new Interval(pre.start, Math.max(pre.end, cur.end));
    			pre = merge;
    		} else {
    			res.add(pre);
    			pre = cur;
    		}
		}
    	res.add(pre);
    	return res;
    }
}