56. Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[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: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

class Solution {
    private class IntervalComparator implements Comparator<int[]> {
        public int compare(int []a, int []b) {
            return a[0] < b[0] ? -1 : a[0] == b[0] ? 0 : 1;
        }
    }
    public int[][] merge(int[][] intervals) {
        int m = intervals.length;
        Collections.sort(Arrays.asList(intervals), new IntervalComparator());
        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()][]);
        
    }
}