LeetCode 435. Non-overlapping Intervals

435. Non-overlapping Intervals

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• Total Accepted: 7406
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• Difficulty: Medium
• Contributors: love_FDU_llp

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.

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【题目分析】

这个题目与《算法导论》中活动安排的题目很是相似。

活动选择问题
有n个须要在同一天使用同一个教室的活动a1,a2,…,an，教室同一时刻只能由一个活动使用。每一个活动ai都有一个开始时间si和结束时间fi 。一旦被选择后，活动ai就占据半开时间区间[si,fi)。若是[si,fi]和[sj,fj]互不重叠，ai和aj两个活动就能够被安排在这一天。该问题就是要安排这些活动使得尽可能多的活动能不冲突的举行。例以下图所示的活动集合S，其中各项活动按照结束时间单调递增排序。

 

考虑使用贪心算法的解法。为了方便，咱们用不一样颜色的线条表明每一个活动，线条的长度就是活动所占据的时间段，蓝色的线条表示咱们已经选择的活动；红色的线条表示咱们没有选择的活动。
若是咱们每次都选择开始时间最先的活动，不能获得最优解：

 

 

若是咱们每次都选择持续时间最短的活动，不能获得最优解：

 

 

能够用数学概括法证实，咱们的贪心策略应该是每次选取结束时间最先的活动。直观上也很好理解，按这种方法选择相容活动为未安排活动留下尽量多的时间。这也是把各项活动按照结束时间单调递增排序的缘由。

【思路】

参照上面活动安排的例子，咱们很容易获得这个题目的解法。这是一个贪心问题，咱们每次都找到那个结束点最小的区间，而后依次向后找那些与前面区间不冲突且结束点早的区间。这个过程当中咱们把局部的最优解合并成了全局的最优解。

【java代码】

 1 /**
 2  * Definition for an interval.
 3  * public class Interval {
 4  *     int start;
 5  *     int end;
 6  *     Interval() { start = 0; end = 0; }
 7  *     Interval(int s, int e) { start = s; end = e; }
 8  * }
 9  */
10 public class Solution {
11     public int eraseOverlapIntervals(Interval[] intervals) {
12         if(intervals.length == 0) return 0;
13         
14         Comparator<Interval> comp = new Comparator<Interval>() {
15             public int compare(Interval interval1, Interval interval2) {
16                 if(interval1.end > interval2.end) return 1;
17                 else if(interval1.end < interval2.end) return -1;
18                 else return 0;
19             }
20         };
21         
22         Arrays.sort(intervals, comp);
23         int lastend = intervals[0].end;
24         int remove = 0;
25         for(int i = 1; i < intervals.length; i++) {
26             if(intervals[i].end == lastend) remove++;
27             else if(intervals[i].start < lastend) remove++;
28             else lastend = intervals[i].end;
29         }
30         
31         return remove;
32     }
33 }