【leetcode刷题笔记】Merge Intervals

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].

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题解：首先对所有的区间按照start大小排序，然后遍历排序后的数组，用last记录前一个区间，如果遍历的当前区间可以和last合并，就把它合并到last里面；否则就把last放到answer list中，并且更新last。

代码如下：

 /**
  * 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 {
     class IntervalComparator implements Comparator<Interval>{
         public int compare(Interval a,Interval b){
             return a.start - b.start;
         }
     }
     public List<Interval> merge(List<Interval> intervals) {
         if(intervals == null || intervals.size() <= 1)
             return intervals;
         Collections.sort(intervals,new IntervalComparator());

         ArrayList<Interval> answer = new ArrayList<Interval>();

         Interval last = intervals.get(0);
         for(int i = 1;i < intervals.size();i++){
             Interval now = intervals.get(i);
             if(last.end >= now.start){
                 last.end = Math.max(last.end, now.end);
             }
             else{
                 answer.add(last);
                 last = now;
             }
         }
         answer.add(last);
         return answer;

     }
 }