LeetCode -- Minimum Number of Arrows to Burst Balloons

题目描述：
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).

用最少数量的箭打完气球。气球的输入以二维数组point[,]表示，其中气球points[i,0]和points[i,1]表示气球横坐标的起始位置。

实现思路：
对气球位置进行排序，从左向右依次打，每次以当前气球最右边的横坐标进行射击。
遍历每个气球，如果当前气球没被打中，更新气球射击位置。

实现代码：

public class Solution {
public int FindMinArrowShots(int[,] points)
{
if(points == null){
return 0;
}
var len = points.GetLength(0);
if(len == 0){
return 0;
}

var balls = new List<Balloon>(len);
for (var i = 0;i < len; i++){
balls.Add(new Balloon(points[i,0], points[i,1]));
}

balls = balls.OrderBy(x=>x.X1).ThenBy(x=>x.X2).ToList();
//Console.WriteLine(balls);
var bulletCount = 1;
var selection = balls.First();
var bulletX = selection.X2;
for (var i = 1;i < balls.Count; i++){
if(balls[i].X1 <= bulletX){
bulletX = Math.Min(bulletX, balls[i].X2);
}else{
bulletX = balls[i].X2;
bulletCount++;
}
}

//Console.WriteLine(balls);

return bulletCount;
}

public class Balloon{
public int X1;
public int X2;
public Balloon(int x1, int x2){
X1 = x1;
X2 = x2;
}
}
}