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], [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 };