标签:
【题目】
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]
.
【analyze】
1.需要先根据每个Interval的start进行排序
2.再根据pre的end是否大于等于cur的start来判断,是否需要进行合并
【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> merge(List<Interval> intervals) { if(intervals==null||intervals.size()<1) return intervals; Collections.sort(intervals,new Comparator<Interval>() { public int compare(Interval i1,Interval i2) { return i1.start-i2.start; }}); List<Interval> res=new ArrayList<Interval>(); 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; } }
标签:
原文地址:http://www.cnblogs.com/hwu2014/p/4479284.html