Given a set of intervals,for each of the interval i,check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i,which can be called that j is on the "right" of i.
For any interval i,you need to store the minimum interval j‘s index,which means that the interval j has the minimum start point to build the "right" relationship for interval i. If the interval j doesn‘t exist,store -1 for the interval i. Finally,you need output the stored value of each interval as an array.
Note:
- You may assume the interval‘s end point is always bigger than its start point.
- You may assume none of these intervals have the same start point.
Example 1:
Input: [ [1,2] ] Output: [-1] Explanation: There is only one interval in the collection,so it outputs -1.
Example 2:
Input: [ [3,4],[2,3],[1,2] ] Output: [-1,1] Explanation: There is no satisfied "right" interval for [3,4]. For [2,the interval [3,4] has minimum-"right" start point; For [1,2],the interval [2,3] has minimum-"right" start point.
Example 3:
Input: [ [1,[3,4] ] Output: [-1,2,-1] Explanation: There is no satisfied "right" interval for [1,4] and [3,4] has minimum-"right" start point.
给定一组区间,对于每一个区间 i,检查是否存在一个区间 j,它的起始点大于或等于区间 i 的终点,这可以称为 j 在 i 的“右侧”。
对于任何区间,你需要存储的满足条件的区间 j 的最小索引,这意味着区间 j 有最小的起始点可以使其成为“右侧”区间。如果区间 j 不存在,则将区间 i 存储为 -1。最后,你需要输出一个值为存储的区间值的数组。
注意:
- 你可以假设区间的终点总是大于它的起始点。
- 你可以假定这些区间都不具有相同的起始点。
示例 1:
输入: [ [1,2] ] 输出: [-1] 解释:集合中只有一个区间,所以输出-1。
示例 2:
输入: [ [3,2] ] 输出: [-1,1] 解释:对于[3,4],没有满足条件的“右侧”区间。 对于[2,3],区间[3,4]具有最小的“右”起点; 对于[1,2],区间[2,3]具有最小的“右”起点。
示例 3:
输入: [ [1,4] ] 输出: [-1,-1] 解释:对于区间[1,4]和[3,4]有最小的“右”起点。
5324ms
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 findRightInterval(_ intervals: [Interval]) -> [Int] { 14 var res:[Int] = [Int]() 15 var v:[Int] = [Int]() 16 var m:[Int:Int] = [Int:Int]() 17 for i in 0..<intervals.count 18 { 19 m[intervals[i].start] = i 20 v.append(intervals[i].start) 21 } 22 v = v.sorted(by:>) 23 for a in intervals 24 { 25 var i:Int = 0 26 while(i < v.count) 27 { 28 if v[i] < a.end 29 { 30 break 31 } 32 i += 1 33 } 34 res.append((i > 0) ? m[v[i - 1]]! : -1) 35 } 36 return res 37 } 38 }
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