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【题目】
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].
【analyze】
1.利用之前类似的一道题目的做法
2.先根据区间的start对区间进行排序
3.再依次插入
【算法】
/** * 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>(); intervals.add(newInterval); Collections.sort(intervals,new Comparator<Interval>() { public int compare(Interval i1,Interval i2) { return i1.start-i2.start; } }); 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; } }
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原文地址：http://www.cnblogs.com/hwu2014/p/4480859.html
