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

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    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:

    1. You may assume the interval's end point is always bigger than its start point.
    2. 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], [1,2], [1,2] ]
    
    Output: 2
    
    Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping. 

    Example 3:

    Input: [ [1,2], [2,3] ]
    
    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:

    1. 可以认为区间的终点总是大于它的起点。
    2. 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: [ [1,2], [2,3], [3,4], [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], [1,2], [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: [ [1,2], [2,3] ]
    
    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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  • 原文地址:https://www.cnblogs.com/strengthen/p/10334487.html
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