Interval List Intersections

Given two lists of closed intervals, each list of intervals is pairwise disjoint and in sorted order.

Return the intersection of these two interval lists.

(Formally, a closed interval [a, b] (with a <= b) denotes the set of real numbers x with a <= x <= b. The intersection of two closed intervals is a set of real numbers that is either empty, or can be represented as a closed interval. For example, the intersection of [1, 3] and [2, 4] is [2, 3].)

Example 1:

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]]

Note:

1. 0 <= A.length < 1000

2. 0 <= B.length < 1000

3. 0 <= A[i].start, A[i].end, B[i].start, B[i].end < 10^9

class Solution {
    public int[][] intervalIntersection(int[][] A, int[][] B) {
        if (A == null || A.length == 0 || B == null || B.length == 0)
            return new int[][] {};

        int m = A.length, n = B.length;
        int i = 0, j = 0;
        List<int[]> res = new ArrayList<>();
        while (i < m && j < n) {
            int[] interval1 = A[i];
            int[] interval2 = B[j];

            // find the overlap... if there is any...
            int startMax = Math.max(interval1[0], interval2[0]);
            int endMin = Math.min(interval1[1], interval2[1]);

            if (endMin >= startMax)
                res.add(new int[] { startMax, endMin });

            // update the pointer with smaller end value...
            if (interval1[1] == endMin)
                i++;
            if (interval2[1] == endMin)
                j++;
        }
        int[][] ans = new int[res.size()][];
        for (i = 0; i < res.size(); i++)
            ans[i] = new int[] { res.get(i)[0], res.get(i)[1] };
        return ans;
    }
}