原题链接在这里：https://leetcode.com/problems/interval-list-intersections/
题目：
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:
0 <= A.length < 10000 <= B.length < 10000 <= A[i].start, A[i].end, B[i].start, B[i].end < 10^9
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
题解：
The example is long, try with a simpler example. Virtualize the result, could see the routine that if the maximum start of two intervals is smaller to or equal with minimum end, then there is an intersection.
Then move the pointer pointering to smaller end.
Time Complexity: O(A.length + B.length).
Space: O(1).
AC Java:
1 class Solution { 2 public int[][] intervalIntersection(int[][] A, int[][] B) { 3 if(A == null || B == null){ 4 return null; 5 } 6 7 List<int []> resList = new ArrayList<>(); 8 int i = 0; 9 int j = 0; 10 while(i<A.length && j<B.length){ 11 int maxStart = Math.max(A[i][0], B[j][0]); 12 int minEnd = Math.min(A[i][1], B[j][1]); 13 if(maxStart <= minEnd){ 14 resList.add(new int[]{maxStart, minEnd}); 15 } 16 17 if(A[i][1]<B[j][1]){ 18 i++; 19 }else{ 20 j++; 21 } 22 } 23 24 int [][] res = new int[resList.size()][2]; 25 for(int ind = 0; ind<resList.size(); ind++){ 26 res[ind] = resList.get(ind); 27 } 28 29 return res; 30 } 31 }