Merge Overlapping Intervals

Given a collection of intervals, merge all overlapping intervals.

For example:

Given [1,3],[2,6],[8,10],[15,18],

return [1,6],[8,10],[15,18].

Make sure the returned intervals are sorted.

思路

我的做法就是先sort，然后从左往右，有overlap就merge。

Solution:

Time: O(nlogn)
Space: O(n)

class Solution {
    public int[][] merge(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> { return Integer.compare(a[0], b[0]); });
        int[][] result = new int[intervals.length][2];
        if (intervals == null || intervals.length == 0) return result;
        int k = 0;
        result[0] = intervals[0];
        for (int i = 1; i < intervals.length; i ++) {
            int[] interval = intervals[i];
            if (result[k][1] < interval[0]) {
                result[++ k] = interval;
            } else {
                result[k][1] = Math.max(result[k][1], interval[1]);
            }
        }
        return Arrays.copyOf(result, k + 1);
    }
}

官方Solution

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public ArrayList<Interval> merge(ArrayList<Interval> intervals) {
        if (intervals == null) return null;
        
        Collections.sort(intervals, (a, b) -> Integer.compare(a.start, b.start));
        ArrayList<Interval> merged = new ArrayList<>();
        
        for (Interval current : intervals) {
            if (merged.isEmpty() || merged.get(merged.size() -1).end < current.start) {
                merged.add(current);
            } else {
                merged.get(merged.size() -1).end = Math.max(current.end, 
                                                   merged.get(merged.size() -1).end);
            }
        }
        return merged;
    }
}