LeetCode-Interval List Intersections

Description：
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]]
Reminder: The inputs and the desired output are lists of Interval objects, and not arrays or lists.

Note:

题意：给定两个序列，每个序列中的元素为一个闭合区间，要求计算这两个序列中所有元素的交集；

解法：因为序列中的元素区间已经按照坐标的大小排好序了，所以我们只需要同时遍历这两个序列，对每两个元素求取他们的区间交集；所以，问题的重点在于如何求解两个区间的交集，我们知道，如果两个区间有交集，那么其开始端一定是两个区间开始端的最大值，结束端一定是两个区间的结束端的最小值，如下如图所示

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
class Solution {
    public Interval[] intervalIntersection(Interval[] A, Interval[] B) {
        List<Interval> res = new ArrayList<>();
        int pa = 0;
        int pb = 0;
        while (pa < A.length && pb < B.length) {
            int st = Math.max(A[pa].start, B[pb].start);
            int ed = Math.min(A[pa].end, B[pb].end);
            if (st <= ed) {
                res.add(new Interval(st, ed));
            }
            if (A[pa].end < B[pb].end) {
                pa++;
            } else {
                pb++;
            }
        }
        
        return res.toArray(new Interval[res.size()]);
    }
}