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.

class Solution {
    public int[][] merge(int[][] intervals) {
        int l = intervals.length;
        if(l == 0 || l == 1) return intervals;
        int[][] res = insert(new int[][]{}, intervals[0]);
            for(int i = 1; i < intervals.length; i++){
                res = insert(res, intervals[i]);
        }
        return res;
    }
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List<int[]> resu = new ArrayList();
        int s = newInterval[0], e = newInterval[1];
        int i = 0;
        //add all intervals which are before the new Interval
        while(i < intervals.length && intervals[i][1] < s){
            resu.add(intervals[i++]);
        }
        //ensure all of the intervals are merged based on the start and end time of the newInterval 
        while(i < intervals.length && intervals[i][0] <= e){
            s = Math.min(s, intervals[i][0]);
            e = Math.max(e, intervals[i][1]);
            i++;
        }
        //Add merged interval
        resu.add(new int[]{s, e});
        //Add the remaining intervals
        while(i < intervals.length) resu.add(intervals[i++]);
        int j = 0;
        int[][] res = new int[resu.size()][2];
        for(int[] tmp: resu){
            res[j++] = tmp;
        }
        return res;
    }
}

使用57题作为method即可，注意parameters