Interval List Intersections
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]]
Note:
0 <= A.length < 1000
0 <= B.length < 1000
0 <= A[i].start, A[i].end, B[i].start, B[i].end < 10^9
class Solution {
public int[][] intervalIntersection(int[][] A, int[][] B) {
if (A == null || A.length == 0 || B == null || B.length == 0)
return new int[][] {};
int m = A.length, n = B.length;
int i = 0, j = 0;
List<int[]> res = new ArrayList<>();
while (i < m && j < n) {
int[] interval1 = A[i];
int[] interval2 = B[j];
// find the overlap... if there is any...
int startMax = Math.max(interval1[0], interval2[0]);
int endMin = Math.min(interval1[1], interval2[1]);
if (endMin >= startMax)
res.add(new int[] { startMax, endMin });
// update the pointer with smaller end value...
if (interval1[1] == endMin)
i++;
if (interval2[1] == endMin)
j++;
}
int[][] ans = new int[res.size()][];
for (i = 0; i < res.size(); i++)
ans[i] = new int[] { res.get(i)[0], res.get(i)[1] };
return ans;
}
}
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