452. Minimum Number of Arrows to Burst Balloons

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of start and end of the diameter suffice. Start is always smaller than end. There will be at most 104 balloons.

An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with xstart and xend bursts by an arrow shot at x if xstart ≤ x ≤ xend. There is no limit to the number of arrows that can be shot. An arrow once shot keeps travelling up infinitely. The problem is to find the minimum number of arrows that must be shot to burst all balloons.

Example:

Input:
[[10,16], [2,8], [1,6], [7,12]]

Output:
2

Explanation:
One way is to shoot one arrow for example at x = 6 (bursting the balloons [2,8] and [1,6]) and another arrow at x = 11 (bursting the other two balloons).

这个题需要考虑多种情况才能做


上边儿两种情况都需要两个arrow。
需要知道什么时候new 一个新的arrow：
如果按照start来排序，我们需要新arrow就是当前面的min(end) < cur.start
图一的a.end < c.start
图二的b.end < c.start

 public int findMinArrowShots(int[][] points) {  
     if(points.length == 0) return 0;  
     int n = points.length;  
     Arrays.sort(points, (a, b)->{  
       if(a[0] == b[0]) return Integer.compare(a[1],b[1]);  
       return Integer.compare(a[0],b[0]);  
     });  
     int p1 = 0;  
     int ans = 1;  
     // point 0's end    
     int end = points[0][1];  
     for(int i = 1; i < n; i++){  
       //when end is smaller than next start, we need another arrow  
       //we need to update end, if end is smaller than previous ones  
       end = Math.min(end,points[i][1]);  
       if( end < points[i][0]){  
         end = points[i][1];  
         ans++;  
       }  
     }  
     return ans;  
   }