Given a collection of intervals,find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
Note:
- You may assume the interval‘s end point is always bigger than its start point.
- Intervals like [1,2] and [2,3] have borders "touching" but they don‘t overlap each other.
Example 1:
Input: [ [1,2],[2,3],[3,4],[1,3] ] Output: 1 Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.
Example 2:
Input: [ [1,2] ] Output: 2 Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
Example 3:
Input: [ [1,3] ] Output: 0 Explanation: You don‘t need to remove any of the intervals since they‘re already non-overlapping.
给定一个区间的集合,找到需要移除区间的最小数量,使剩余区间互不重叠。
注意:
- 可以认为区间的终点总是大于它的起点。
- 区间 [1,2] 和 [2,3] 的边界相互“接触”,但没有相互重叠。
示例 1:
输入: [ [1,3] ] 输出: 1 解释: 移除 [1,3] 后,剩下的区间没有重叠。
示例 2:
输入: [ [1,2] ] 输出: 2 解释: 你需要移除两个 [1,2] 来使剩下的区间没有重叠。
示例 3:
输入: [ [1,3] ] 输出: 0 解释: 你不需要移除任何区间,因为它们已经是无重叠的了。
76ms
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 eraSEOverlapIntervals(_ intervals: [Interval]) -> Int { 14 var intervals = intervals 15 if intervals.isEmpty {return 0} 16 intervals.sort(by:{(_ a:Interval,_ b:Interval) -> Bool in return a.start < b.start}) 17 var res:Int = 0 18 var n:Int = intervals.count 19 var endLast:Int = intervals[0].end 20 for i in 1..<n 21 { 22 var t:Int = endLast > intervals[i].start ? 1 : 0 23 endLast = t == 1 ? min(endLast,intervals[i].end) : intervals[i].end 24 res += t 25 } 26 return res 27 } 28 }
80ms
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 eraSEOverlapIntervals(_ intervals: [Interval]) -> Int { 14 if intervals.count <= 1 { 15 return 0 16 } 17 18 var move = 1 19 var ts = intervals 20 ts.sort { (i1,i2) -> Bool in 21 return i1.end <= i2.end 22 } 23 24 var temp = ts[0] 25 for i in 1..<ts.count { 26 let start = ts[i].start 27 if start >= temp.end { 28 move += 1 29 temp = ts[i] 30 } 31 } 32 return intervals.count - move 33 } 34 }
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