You are given two lists of closed intervals, firstList
and secondList
, where firstList[i] = [starti, endi]
and secondList[j] = [startj, endj]
. Each list of intervals is pairwise disjoint and in sorted order.
Return the intersection of these two interval lists.
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 are either empty or represented as a closed interval. For example, the intersection of [1, 3]
and [2, 4]
is [2, 3]
.
Example 1:
Input: firstList = [[0,2],[5,10],[13,23],[24,25]], secondList = [[1,5],[8,12],[15,24],[25,26]] Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
Example 2:
Input: firstList = [[1,3],[5,9]], secondList = [] Output: []
Example 3:
Input: firstList = [], secondList = [[4,8],[10,12]] Output: []
Example 4:
Input: firstList = [[1,7]], secondList = [[3,10]] Output: [[3,7]]
Constraints:
0 <= firstList.length, secondList.length <= 1000
firstList.length + secondList.length >= 1
0 <= starti < endi <= 109
endi < starti+1
0 <= startj < endj <= 109
endj < startj+1
Companies:
Facebook, Uber, Yandex, Google, Amazon
Related Topics:
Array, Two Pointers
Similar Questions:
// OJ: https://leetcode.com/problems/interval-list-intersections/
// Author: github.com/lzl124631x
// Time: O(M+N)
// Space: O(1)
class Solution {
public:
vector<vector<int>> intervalIntersection(vector<vector<int>>& A, vector<vector<int>>& B) {
int M = A.size(), N = B.size();
vector<vector<int>> ans;
for (int i = 0, j = 0; i < M && j < N; ) {
int s = max(A[i][0], B[j][0]), e = min(A[i][1], B[j][1]);
if (s <= e) ans.push_back({ s, e });
if (A[i][1] < B[j][1]) ++i;
else ++j;
}
return ans;
}
};