435. Non-overlapping Intervals

Posted 2021-02-06 __Meng

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.


求移除多少区间后，剩余区间都是不重叠的。

先求最多能组成多少不重叠的区间，再用总区间数减去不重叠区间数。

C++：

 1 /**
 2  * Definition for an interval.
 3  * struct 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 bool compare(const Interval& a ,const Interval& b){
11         return a.end < b.end ;
12 }
13 class Solution {
14 public:
15     int eraseOverlapIntervals(vector<Interval>& intervals) {
16         if (intervals.size() == 0){
17             return 0 ;
18         }
19         sort(intervals.begin() , intervals.end() , compare) ;
20         int cnt = 1; 
21         int end = intervals[0].end ;
22         for(int i = 1 ; i < intervals.size() ; i++){
23             if (intervals[i].start < end){
24                 continue ;
25             }
26             end = intervals[i].end ;
27             cnt++ ;
28         }
29         return intervals.size() - cnt ;
30     }
31 };