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] 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 {
fun merge(intervals: Array<IntArray>): Array<IntArray> {
val result: MutableList<IntArray> = mutableListOf()
if (intervals.size <= 1) return intervals
intervals.sortBy { it.first() }
var currentRange = 1
var begin = intervals[0][0]
var end = intervals[0][1]
while (currentRange < intervals.size) {
if (intervals[currentRange][0] > end) {
result.add(intArrayOf(begin, end))
begin = intervals[currentRange][0]
end = intervals[currentRange][1]
} else if (intervals[currentRange][1] > end) {
end = intervals[currentRange][1]
}
currentRange++
}
result.add(intArrayOf(begin, end))
return result.toTypedArray()
}
}