Topic38——56. Consolidation interval

subject ： In array intervals Represents a set of intervals , The single interval is intervals[i] = [starti, endi] . Please merge all overlapping intervals , And back to An array of non overlapping intervals , The array needs to cover exactly 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]]
 explain ： Section  [1,3]  and  [2,6]  overlap ,  Combine them into  [1,6].

 Example  2：
 Input ：intervals = [[1,4],[4,5]]
 Output ：[[1,5]]
 explain ： Section  [1,4]  and  [4,5]  Can be regarded as overlapping interval .

Tips ：
1 <= intervals.length <= 104
intervals[i].length == 2
0 <= starti <= endi <= 104

class Solution {
    
    public int[][] merge(int[][] intervals) {
    
        // Sort two-dimensional arrays 
        Arrays.sort(intervals, new Comparator<int[]>() {
    
            public int compare(int[] a, int[] b) {
    
                if(a[0] < b[0] || (a[0] == b[0] && a[1] < b[1])) {
    
                    return -1;
                } else if(a[0] == b[0] && a[1] == b[1]) {
    
                    return 0;
                } else {
    
                    return 1;
                }
            }
        });
        List<int[]> newArr = new ArrayList<>();
        newArr.add(intervals[0]);
        int index = 0;
        for(int i = 1; i < intervals.length; i++) {
    
            if(intervals[i][0] <= newArr.get(index)[1]) {
    
                newArr.get(index)[1] = Math.max(intervals[i][1], newArr.get(index)[1]); 
            } else {
    
                newArr.add(intervals[i]);
                index++;
            }
        }
        return newArr.toArray(new int[newArr.size()][]);
    }
}