Day-15-Non-overlapping Intervals.cpp

Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.



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.


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.
















class Solution {
public:

    static bool comp(vector<int> a, vector<int> b){
        return a[1] < b[1];
    }

    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        int res=0;
        int n=intervals.size();
        if(n==0) return 0;

        sort(intervals.begin(), intervals.end(), comp);


        pair<int, int> cur;
        cur.first=intervals[0][0];
        cur.second=intervals[0][1];

        for(int i=1;i<n;i++){
            if(cur.second>intervals[i][0]){
                res++;
            } else {
                cur.first=intervals[i][0];
                cur.second=intervals[i][1];
            }
        }
        return res;
    }
};