57.
1. Question
Given a set ofnon-overlappingintervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1:
Given intervals[1,3],[6,9], insert and merge[2,5]in as[1,5],[6,9].
Example 2:
Given[1,2],[3,5],[6,7],[8,10],[12,16], insert and merge[4,9]in as[1,2],[3,10],[12,16].
This is because the new interval[4,9]overlaps with[3,5],[6,7],[8,10].
2. Implementation
(1) Scan Line
/**
* 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 {
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
List<Interval> res = new ArrayList<>();
for (Interval interval : intervals) {
if (newInterval == null || interval.end < newInterval.start) {
res.add(interval);
}
else if (newInterval.end < interval.start) {
res.add(newInterval);
res.add(interval);
newInterval = null;
}
else {
newInterval.start = Math.min(interval.start, newInterval.start);
newInterval.end = Math.max(interval.end, newInterval.end);
}
}
if (newInterval != null) {
res.add(newInterval);
}
return res;
}
}
3. Time & Space Complexity
Scan Line: 时间复杂度O(n), 空间复杂度O(n)