# 452. Minimum Number of Arrows to Burst Balloons

## Problem

There are some 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. The start is always smaller than the end.

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 traveling up infinitely.

Given an array points where points[i] = [xstart, xend], return the minimum number of arrows that must be shot to burst all balloons.

Example 1:

```
Input: points = [[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).
```

Example 2:

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

Example 3:

```
Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2
```

## Code

```
class Solution {
public int findMinArrowShots(int[][] points) {
Arrays.sort(points, (a, b) -> {
// 不能使用 a[0] - b[0]
// 数字太大会造成overflow
// [[-2147483646,-2147483645],[2147483646,2147483647]]
if (a[0] == b[0]) return 0;
if (a[0] < b[0]) return -1;
return 1;
});
int res = 1;
int[] start = points[0];
for(int i = 1; i < points.length; i++) {
int[] curr = points[i];
if(curr[0] > start[1]) {
res++;
start = curr;
} else {
start[0] = Math.max(start[0], curr[0]);
start[1] = Math.min(start[1], curr[1]);
}
}
return res;
}
}
```