【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;
        
         
    }
}