【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：

 1 /**

 2  * Definition for an interval.  3  * public class Interval {  4  * int start;  5  * int end;  6  * Interval() { start = 0; end = 0; }  7  * Interval(int s, int e) { start = s; end = e; }  8  * }  9  */

10 public class Solution { 11     class IntervalComparator implements Comparator<Interval>{ 12         public int compare(Interval a,Interval b){ 13             return a.start - b.start; 14  } 15  } 16     public List<Interval> merge(List<Interval> intervals) { 17         if(intervals == null || intervals.size() <= 1) 18             return intervals; 19         Collections.sort(intervals,new IntervalComparator()); 20         

21         ArrayList<Interval> answer = new ArrayList<Interval>(); 22         

23         Interval last = intervals.get(0); 24         for(int i = 1;i < intervals.size();i++){ 25             Interval now = intervals.get(i); 26             if(last.end >= now.start){ 27                 last.end = Math.max(last.end, now.end); 28  } 29             else{ 30  answer.add(last); 31                 last = now; 32  } 33  } 34  answer.add(last); 35         return answer; 36         

37          

38  } 39 }