Hà Phan（河）

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]]
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.

## Solution

Because two arrays are sorted, we can do a straight check.
Compare one in A by one in B, move to next element when current element's head is lower.

``````class Solution {
func intervalIntersection(_ A: [[Int]], _ B: [[Int]]) -> [[Int]] {
var moving: Int = 0

var indexA = 0
var indexB = 0

var res = [[Int]]()

while (indexA < A.count) && (indexB < B.count) {
let valA = A[indexA]
let valB = B[indexB]

if let intersection = valA.intersectionWith(valB) {
res.append(intersection)
}

if valA > valB {
indexB += 1
}else{
indexA += 1
}
}

return res
}
}

extension Array where Element == Int {
func intersectionWith(_ range: [Int]) -> [Int]?{
if self < range || self > range {
return nil
}
return [self >= range ? self : range, self <= range ? self : range]
}
}
``````