[Swift]LeetCode435. 无重叠区间 | Non-overlapping Intervals

发布时间：2020-12-14 05:05:12 所属栏目：百科来源：网络整理

导读：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

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 }

（编辑：李大同）

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