1232. Minimum number of arrows for exploding balloons

describe

There are many two-dimensional balloons . For each balloon , The input provided is the start and end coordinates of the horizontal diameter . Because it's horizontal , therefore y The coordinates don't matter , Therefore, the starting point and ending point of the diameter x Coordinates are enough . The starting point is always less than the ending point . There will be at most 10^4 A balloon .

You can go along x The axis emits arrows vertically upward from different points . If xstart≤x≤xend, Then the coordinates are xstart and xend The balloon was in x The arrow shot at . There is no limit to the number of arrows that can be fired . The arrow of a shot keeps moving up infinitely . The problem is to find the minimum number of launch arrows that pierce all balloons .

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Examples 1

 Input :
[[10,16], [2,8], [1,6], [7,12]]
 Output :
2
 explain ：
 One way is to fire an arrow , For example, in x = 6（ Blasting balloon [2,8] and [1,6]）, Launch another arrow at x = 11（ Blow up the other two balloons ）.

Examples 2

 Input :
[[1,2],[3,4],[5,6],[7,8]]
 Output :
4

 

    int findMinArrowShots(vector<vector<int>> &points) {

        // Write your code here

        if(0 == points.size())

        {

            return 0;

        }

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

        int count=1;

        int x=points[0][1];

        for(int i=1;i<points.size();i++)

        {

            if(x < points[i][0])

            {

                x= points[i][1];

                count++;

            }

        }

        return count;

 

    }