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.
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].
Solution：Interative
思路：
Screen Shot 2017-11-21 at 12.56.17.png
Time Complexity: O(N) Space Complexity: O(1)
Solution Code：
class Solution {
public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
List<Interval> result = new ArrayList<>();
int i = 0;
// add all the intervals ending before newInterval starts
while(i < intervals.size() && newInterval.start > intervals.get(i).end) {
result.add(intervals.get(i));
i++;
}
int new_start = newInterval.start;
int new_end = newInterval.end;
// merge all overlapping intervals to one considering newInterval
while(i < intervals.size() && newInterval.end >= intervals.get(i).start) {
new_start = Math.min(new_start, intervals.get(i).start);
new_end = Math.max(new_end, intervals.get(i).end);
i++;
}
result.add(new Interval(new_start, new_end)); // add the union of intervals we got
// add all the rest
while (i < intervals.size()) {
result.add(intervals.get(i));
i++;
}
return result;
}
}
