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>(); int index = 0; while (index < intervals.size() && intervals.get(index).end < newInterval.start) { res.add(intervals.get(index++)); } while (index < intervals.size() && intervals.get(index).start <= newInterval.end){ newInterval = new Interval(Math.min(newInterval.start, intervals.get(index).start), Math.max(newInterval.end, intervals.get(index).end)); index++; } res.add(newInterval); while (index < intervals.size()) res.add(intervals.get(index++)); return res; }}新闻热点
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