Leetcode 757 set the intersection size to at least 2[sort greedy] the leetcode path of heroding

Their thinking ：
A relatively complex greedy problem , Because at least two intersecting elements , Therefore, we need to define an array for each interval to store intersecting intervals to reduce the number of repetitions and ensure the minimum output . First, sort the array according to the beginning of the interval , Then traverse the array from back to front , For each interval , Choose several numbers according to the number that coincides with the interval of thick surface （ Of course not more than 2）, If an interval coincides , Just choose another one , Overlap the two intervals and skip directly , The code is as follows ：

class Solution {
    
public:
    void help(vector<vector<int>>& intervals, vector<vector<int>>& temp, int pos, int num) {
    
        for (int i = pos; i >= 0; i--) {
    
            if (intervals[i][1] < num) {
    
                break;
            }
            temp[i].push_back(num);
        }
    }

    int intersectionSizeTwo(vector<vector<int>>& intervals) {
    
        int n = intervals.size();
        int res = 0;
        sort(intervals.begin(), intervals.end(), [&](vector<int>& a, vector<int>& b) {
    
            if (a[0] == b[0]) {
    
                return a[1] > b[1];
            }
            return a[0] < b[0];
        });
        vector<vector<int>> temp(n);
        for (int i = n - 1; i >= 0; i--) {
    
            for (int j = intervals[i][0], k = temp[i].size(); k < 2; j++, k++) {
    
                res++;
                help(intervals, temp, i - 1, j);
            }
        }
        return res;
    }
};