Medium
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] overlap, 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 {
List<List<int>> merge(List<List<int>> intervals) {
// Sort the intervals based on the start time
intervals.sort((a, b) =>a[0].compareTo(b[0]));
List<List<int>> merged = [];
List<int> current = intervals[0];
merged.add(current);
for (List<int> next in intervals){
if (current[1] >= next[0]) {
// If the intervals overlap, merge them
current[1] = current[1] > next[1] ? current[1] : next[1];
} else {
// Otherwise, move to the next interval
current = next;
merged.add(current);
}
}
return merged;
}
}