Given a collection of intervals, merge all overlapping intervals.
Example 1:
Input: [[1,3],[2,6],[8,10],[15,18]] Output: [[1,6],[8,10],[15,18]] Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
Example 2:
Input: [[1,4],[4,5]] Output: [[1,5]] Explanation: Intervals [1,4] and [4,5] are considered overlapping.
方法1：
/**
* 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; }
* }
*/
class Solution {
private class IntervalComparator implements Comparator<Interval> {
@Override
public int compare(Interval a, Interval b) {
//return a.start < b.start ? -1 : a.start == b.start ? 0 : 1;
if(a.start<b.start){
return -1;
}else{
if(a.start==b.start){
return 0;
}else{
return 1;
}
}
}
}
public List<Interval> merge(List<Interval> intervals) {
Collections.sort(intervals, new IntervalComparator());
for(Interval interval:intervals){
System.out.println(interval.start+"--"+interval.end);
}
LinkedList<Interval> merged = new LinkedList<Interval>();
for (Interval interval : intervals) {
// if the list of merged intervals is empty or if the current
// interval does not overlap with the previous, simply append it.
if (merged.isEmpty() || merged.getLast().end < interval.start) {
merged.add(interval);
}
// otherwise, there is overlap, so we merge the current and previous
// intervals.
else {
merged.getLast().end = Math.max(merged.getLast().end, interval.end);
}
}
return merged;
}
}时间复杂度：O(n.logn)
空间复杂度：O(n)