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.
Note:
0 <= A.length < 1000
0 <= B.length < 1000
0 <= A[i].start,A[i].end,B[i].start,B[i].end < 10^9
给定两个由一些闭区间组成的列表,每个区间列表都是成对不相交的,并且已经排序。
返回这两个区间列表的交集。
(形式上,闭区间 [a,b]
(其中 a <= b
)表示实数 x
的集合,而 a <= x <= b
。两个闭区间的交集是一组实数,要么为空集,要么为闭区间。例如,[1,3] 和 [2,4] 的交集为 [2,3]。)
示例:
输入:A = [[0,B = [[1,26]] 输出:[[1,25]] 注意:输入和所需的输出都是区间对象组成的列表,而不是数组或列表。
提示:
0 <= A.length < 1000
0 <= B.length < 1000
0 <= A[i].start,B[i].end < 10^9
1472 ms
1 /** 2 * DeFinition for an interval. 3 * public class Interval { 4 * public var start: Int 5 * public var end: Int 6 * public init(_ start: Int,_ end: Int) { 7 * self.start = start 8 * self.end = end 9 * } 10 * } 11 */ 12 class Solution { 13 func intervalIntersection(_ A: [Interval],_ B: [Interval]) -> [Interval] { 14 if A.isEmpty || B.isEmpty {return []} 15 var C:[Interval] = [Interval](repeating:Interval(0,1),count:A.count*B.count) 16 var p:Int = 0 17 for x in A 18 { 19 for y in B 20 { 21 var l:Int = max(x.start,y.start) 22 var r:Int = min(x.end,y.end) 23 if l <= r 24 { 25 C[p] = Interval(l,r) 26 p += 1 27 } 28 } 29 } 30 if (p - 1) < C.count 31 { 32 var arr = C[0...(p - 1)] 33 return [Interval](arr) 34 } 35 return C 36 } 37 }
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