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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: [[1,25]] Reminder: The inputs and the desired output are lists of Interval objects,and not arrays or lists.
1 class@H_502_49@ Solution { 2 public int[][] intervalIntersection(int[][] A,int@H_502_49@[][] B) { 3 if (A == null || A.length == 0 || B == null || B.length == 0@H_502_49@) { 4 return new int[0][0@H_502_49@]; 5 @H_502_49@ } 6 7 int m = A.length,n =@H_502_49@ B.length; 8 int i = 0,j = 0@H_502_49@; 9 List<int[]> res = new ArrayList<>@H_502_49@(); 10 while (i < m && j <@H_502_49@ n) { 11 int[] a =@H_502_49@ A[i]; 12 int[] b =@H_502_49@ B[j]; 13 14 // find the overlap... if there is any... 15 int startMax = Math.max(a[0],b[0@H_502_49@]); 16 int endMin = Math.min(a[1],b[1@H_502_49@]); 17 18 if (endMin >=@H_502_49@ startMaX) { 19 res.add(new int@H_502_49@[]{startMax,endMin}); 20 @H_502_49@ } 21 22 //update the pointer with @R_696_11412@ller end value... 23 if (a[1] == endMin) { i++@H_502_49@; } 24 if (b[1] == endMin) { j++@H_502_49@; } 25 @H_502_49@ } 26 return res.toArray(new int[0][0@H_502_49@]); 27 @H_502_49@ } 28 }
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