题目难度:Hard
Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]
Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
/**
* 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> result = new LinkedList<>();
int i = 0;
// add all the intervals ending before newInterval starts
while (i < intervals.size() && intervals.get(i).end < newInterval.start)
result.add(intervals.get(i++));
// merge all overlapping intervals to one considering newInterval
while (i < intervals.size() && intervals.get(i).start <= newInterval.end) {
newInterval = new Interval( // we could mutate newInterval here also
Math.min(newInterval.start, intervals.get(i).start),
Math.max(newInterval.end, intervals.get(i).end));
i++;
}
result.add(newInterval); // add the union of intervals we got
// add all the rest
while (i < intervals.size()) result.add(intervals.get(i++));
return result;
}
}