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

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).

對於有部分重疊的區間（線段），每次只保留一個，留下一個右端點“最靠右”的區間（線段），對刪除的區間（線段）進行計數，總共刪除了K個區間，最後的答案等於points.length()-K

public class Solution {
    public static int findMinArrowShots(int[][] points)
	{
		int m=points.length;
		if(m<1)
			return 0;
		
		Arrays.sort(points,new Comparator<int[]>()
		{
			@Override
			public int compare(int[] o1, int[] o2)
			{
				// TODO Auto-generated method stub
				if(o1[0]!=o2[0])
					return o1[0]-o2[0];
				return o1[1]-o2[1];
			}
		});
		
		int ret=0;
		int prev=points[0][1];
		for(int i=1;i<m;i++)
		{
			if(points[i][0]<=prev)
			{
				ret++;
				if(prev<=points[i][1])
					continue;
			}
			
			prev=points[i][1];
		}
		
		return m-ret;
	}
}