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 < 1000
0 <= B.length < 1000
0 <= 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.
代码
Approach #1 Merge Intervals
Time && Space Complexity: O(M + N)
classSolution {publicint[][] intervalIntersection(int[][] A,int[][] B) {List<int[]> ans =newArrayList();int i =0, j =0;while (i <A.length&& j <B.length) {int lo =Math.max(A[i][0],B[j][0]);int hi =Math.min(A[i][1],B[j][1]);if (lo <= hi) {ans.add(newint[]{lo, hi}); }// 合并小的endif (A[i][1] <B[j][1]) { i++; } else { j++; } }returnans.toArray(newint[ans.size()][]); }}