# 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:**

![](https://assets.leetcode.com/uploads/2019/01/30/interval1.png)

<pre><code><strong>Input: firstList = [[0,2],[5,10],[13,23],[24,25]], secondList = [[1,5],[8,12],[15,24],[25,26]]
</strong><strong>Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: firstList = [[1,3],[5,9]], secondList = []
</strong><strong>Output: []
</strong></code></pre>

**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()][]);
}
```
