leetcode435. Non overlapping interval

One ： The analects of Confucius

different No common goal, no common cause to cooperate We have no reason to hold our own judgment Look at what others have experienced , There is no real empathy when you start again , What we can do is to respect the views of others , Keep your attitude .

Two ： subject

3、 ... and : Upper code

class Solution {
    
public:
    /**  Ideas :1. Here, we still handle it in ascending order according to the second element of the interval array  [[1,2],[1,3],[2,3],[3,4]]  Here, as long as the right boundary of our interval array is less than the left boundary of the next array   Then our final interval is   Non overlapping . */    
    static bool cmp(const vector<int> & v1,const vector<int>& v2) {
    
        return v1[1] < v2[1];
    }

    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
    

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

        int temp = intervals[0][1];// First right boundary 
        int count = 1;// The number of interval arrays that meet the requirements （ Record first   An interval array ）
    
        for(int i = 0; i < intervals.size(); i++) {
    
            if(temp <= intervals[i][0]){
    // If it is smaller than the left boundary , Then update the right boundary , At the same time, record the number of interval arrays that meet the requirements 
                temp = intervals[i][1];
                count++;
            }
        }

        return intervals.size() - count;

    }
};

come on. Good night, A stranger