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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].
与上一题基本一样的解法。
public class Solution { public List<Interval> insert(List<Interval> intervals, Interval newInterval) { List<Interval> res = new ArrayList<Interval>(); intervals.add(newInterval); res = merge(intervals); return res; } public List<Interval> merge(List<Interval> intervals) { if (intervals == null || intervals.size() <= 1) { return intervals; } Collections.sort(intervals, new IntervalComparator()); List<Interval> res = new ArrayList<Interval>(); int i=0; while(i<intervals.size()) { Interval newint = new Interval(); int newstart = intervals.get(i).start; int newend = intervals.get(i).end; int j=i+1; while(j<intervals.size()&&newend>=intervals.get(j).start) { newend = Math.max(newend,intervals.get(j).end); j++; } newint.start=newstart; newint.end=newend; res.add(newint); if(j!=i+1) i=j; else i++; } return res; } private class IntervalComparator implements Comparator<Interval> { public int compare(Interval a, Interval b) { return a.start - b.start; } } }
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原文地址：http://www.cnblogs.com/hygeia/p/5112649.html
