模板题目：Interval/Mergelist （986 Interval List Intersections/56 Merge Intervals）

这两道题目 简直就是怪异方法使用大合集 toArray(), copyOf()从来不敢使用 现在 都用上了 现在是时候好好研究一下这几个方法了（其实这种题目比较恶心的点就在于这个容器的转换）
986

Input: A = [[0,2],[5,10],[13,23],[24,25]], B = [[1,5],[8,12],[15,24],[25,26]]
Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]

class Solution {
  public int[][] intervalIntersection(int[][] A, int[][] B) {
    List<int[]> ans = new ArrayList();//外层是List 内层是int[] 就是指不知道有多少个int[]对
    int i = 0, j = 0;

    while (i < A.length && j < B.length) {
      // Let's check if A[i] intersects B[j].
      // lo - the startpoint of the intersection
      // hi - the endpoint of the intersection
      int lo = Math.max(A[i][0], B[j][0]); //起点的最大
      int hi = Math.min(A[i][1], B[j][1]); //终点的最小
      //此时的low high就是可能的交集 
      if (lo <= hi) //
        ans.add(new int[]{lo, hi});

      // Remove the interval with the smallest endpoint
      if (A[i][1] < B[j][1])
        i++;
      else
        j++;
    }

    return ans.toArray(new int[ans.size()][]);
  }
}

56
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].

class Solution {
    public int[][] merge(int[][] intervals) {
        if (intervals == null || intervals.length <= 1) {
            return intervals;
        }
        List<int[]> res = new ArrayList<>();
        
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        res.add(intervals[0]);
        for (int i = 1; i != intervals.length; i++) {
            if (res.get(res.size() - 1)[1] < intervals[i][0]) {
                res.add(intervals[i]);
            } else {
                int left = Math.min(res.get(res.size() - 1)[0], intervals[i][0]); 
                int right = Math.max(res.get(res.size() - 1)[1], intervals[i][1]);
                res.get(res.size() - 1)[0] = left;
                res.get(res.size() - 1)[1] = right;
            }
        }
        return res.toArray(new int[res.size()][]);
    }
}