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
function merge(intervals: number[][]): number[][] {
let result = []
intervals.sort((a, b) => a[0] - b[0])
for (let i = 0; i < intervals.length; i++) { //NOSONAR
let len = result.length
if (len > 0 && intervals[i][0] <= result[len - 1][1]) {
if (result[len - 1][0] >= intervals[i][0]) {
result[len - 1][0] = intervals[i][0]
}
if (result[len - 1][1] <= intervals[i][1]) {
result[len - 1][1] = intervals[i][1]
}
} else {
result.push(intervals[i])
}
}
return result
}
export { merge }