986. Interval List Intersections

986。Interval List Intersections

You are given two lists of closed intervals, firstList and secondList, where firstList[i] = [starti, endi] and secondList[j] = [startj, endj]. Each list of intervals is pairwise disjoint and in sorted order.

Return the intersection of these two interval lists.

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 are either empty or represented as a closed interval. For example, the intersection of [1, 3] and [2, 4] is [2, 3].

Example 1:

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

Example 2:

Input: firstList = [[1,3],[5,9]], secondList = []
Output: []

Constraints:

• 0 <= firstList.length, secondList.length <= 1000

• firstList.length + secondList.length >= 1

• 0 <= starti < endi <= 109

• endi < starti+1

• 0 <= startj < endj <= 109

• endj < startj+1

My Solutions:

public int[][] intervalIntersection(int[][] firstList, int[][] secondList) {
    int firstLen = firstList.length, secondLen = secondList.length;
    if (firstLen == 0 || secondLen == 0) return new int[][]{};

    List<int[]> list = new ArrayList<>();
    int i = 0, j = 0; 
    while (i < firstLen && j < secondLen) {
        int latestStart = Math.max(firstList[i][0], secondList[j][0]);
        int earliestEnd = Math.min(firstList[i][1], secondList[j][1]);
        if (latestStart <= earliestEnd) list.add(new int[]{latestStart, earliestEnd});

        if (firstList[i][1] < secondList[j][1]) i++; // first is in front of second
        else j++; // second is in front of first
    }
    return list.toArray(new int[list.size()][]);
}