leetcode452. Detonate the balloon with the minimum number of arrows

One : The analects of Confucius

Less self touching , Thank you for your reflection , Always be alert whether you are pretending to work hard , Do you understand Will you What do you want Only myself The clearest

Two ： subject

3、 ... and : Upper code

class Solution {
    
public:
    /**  Ideas : 1. This problem is similar to non overlapping space , Let's deal with it in ascending order according to the right boundary of each balloon  2. Then we begin to compare the right boundary of the balloon with the left boundary of the next balloon , If it is bigger than it, it can be blasted with the same arrow   If it's smaller than that, update the boundary   At the same time, the number of bows and arrows is increased by one . */

    static bool cmp(const vector<int>& v1,const vector<int>& v2) {
    
        return v1[1] < v2[1];
    }

    int findMinArrowShots(vector<vector<int>>& points) {
    
        
        sort(points.begin(),points.end(),cmp);
        int temp = points[0][1];// The first right boundary 
        int count = 1;// The initial value is 1 It means that we need a bow and arrow for the first time 

        for(int i = 0; i < points.size(); i++) {
    

            if(temp < points[i][0]) {
    // If the right boundary of the balloon is smaller than the left boundary of the next balloon , Then add one to the number of bows and arrows 
                temp = points[i][1];// Also update the right boundary 
                count++;
            }
        }

        return count;
    }
};