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.
package leetcode;
import java.util.Arrays;
import java.util.Comparator;
import java.util.LinkedList;
/**
* Created by mkhwang on 2021/05/13
* Email : orange2652@gmail.com
* Github : https://github.com/myeongkwonhwang
*/
public class MergeIntervals {
static class Solution {
public int[][] merge(int[][] intervals) {
Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
LinkedList<int[]> merged = new LinkedList<>();
for (int[] interval : intervals) {
if (merged.isEmpty() || merged.getLast()[1] < interval[0]) {
merged.add(interval);
} else {
merged.getLast()[1] = Math.max(merged.getLast()[1], interval[1]);
}
}
return merged.toArray(new int[merged.size()][]);
}
}
}