问题描述
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].
code
/*
先排序吧Arrays【数组[]】Collections【链表list】
再一个一个比较吧
*/
/**
* 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.size() < 2) return intervals;
Collections.sort(intervals,new MyComparator());
Iterator<Interval> it = intervals.iterator();
Interval now = it.next();
while(it.hasNext()) {
Interval next = it.next();
if(isOverlap(now, next)) {
now.start = Math.min(now.start,next.start);
now.end = Math.max(now.end,next.end);
it.remove();
} else if(isWithin(now,next))
it.remove();
else {
now = next;
}
}
return intervals;
}
private boolean isOverlap(Interval a, Interval b) {
return a.end >= b.start;
}
private boolean isWithin(Interval a, Interval b) {
return a.start <= b.start && a.end >= b.end;
}
private class MyComparator implements Comparator<Interval> {
public int compare(Interval a,Interval b) {
return a.start-b.start;
}
}
}