Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.
Example 1:
Input: intervals = [[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: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.
Constraints:
- 1 <= intervals.length <= 104
- intervals[i].length == 2
- 0 <= starti <= endi <= 104
class Solution {
public int[][] merge(int[][] intervals) {
if (intervals.length <= 1)
return intervals;
// Sort by ascending starting point
Arrays.sort(intervals, (i1, i2) -> Integer.compare(i1[0], i2[0]));
List<int[]> result = new ArrayList<>();
int[] newInterval = intervals[0];
result.add(newInterval);
for (int[] interval : intervals) {
if (interval[0] <= newInterval[1]) // Overlapping intervals, move the end if needed
newInterval[1] = Math.max(newInterval[1], interval[1]);
else { // Disjoint intervals, add the new interval to the list
newInterval = interval;
result.add(newInterval);
}
}
return result.toArray(new int[result.size()][]);
}
}