Insert Interval

最新推荐文章于 2020-09-09 23:29:23 发布

最新推荐文章于 2020-09-09 23:29:23 发布 · 69 阅读

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本文介绍了一种在已排序的不重叠区间中插入新区间并进行必要合并的方法。通过两个示例说明了如何处理区间重叠的情况，最终实现区间列表的更新。

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

/**
 * 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> insert(List<Interval> intervals, Interval newInterval) {
    	List<Interval> res = new ArrayList<Interval>();
    	for (Interval in : intervals) {
    		if (newInterval.start > in.end) {
    			res.add(in);
    		} else if (newInterval.end < in.start) {
    			res.add(newInterval);
    			newInterval = in;
    		} else if (newInterval.start<=in.end || newInterval.end>=in.start) {
    			int start = Math.min(newInterval.start, in.start);
    			int end = Math.max(in.end, newInterval.end);
    			newInterval = new Interval(start, end);
    		}
    	}
    	res.add(newInterval);
    	return res;
    }
}