435. Non-overlapping Intervals

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++：

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
bool compare(const Interval& a ,const Interval& b){
        return a.end < b.end ;
}
class Solution {
public:
    int eraseOverlapIntervals(vector<Interval>& intervals) {
        if (intervals.size() == 0){
            return 0 ;
        }
        sort(intervals.begin() , intervals.end() , compare) ;
        int cnt = 1; 
        int end = intervals[0].end ;
        for(int i = 1 ; i < intervals.size() ; i++){
            if (intervals[i].start < end){
                continue ;
            }
            end = intervals[i].end ;
            cnt++ ;
        }
        return intervals.size() - cnt ;
    }
};