# 452. Minimum Number of Arrows to Burst Balloons

Medium
There are some spherical balloons taped onto a flat wall that represents the XY-plane. The balloons are represented as a 2D integer array `points` where `points[i] = [xstart, xend]` denotes a balloon whose horizontal diameter stretches between `xstart` and `xend`. You do not know the exact y-coordinates of the balloons.
Arrows can be shot up directly vertically (in the positive y-direction) from different points along the x-axis. A balloon with `xstart` and `xend` is burst by an arrow shot at `x` if `xstart <= x <= xend`. There is no limit to the number of arrows that can be shot. A shot arrow keeps traveling up infinitely, bursting any balloons in its path.
Given the array `points`, 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: The balloons can be burst by 2 arrows:
- Shoot an arrow at x = 6, bursting the balloons [2,8] and [1,6].
- Shoot an arrow at x = 11, bursting the balloons [10,16] and [7,12].
Example 2:
Input: points = [[1,2],[3,4],[5,6],[7,8]]
Output: 4
Explanation: One arrow needs to be shot for each balloon for a total of 4 arrows.
Example 3:
Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2
Explanation: The balloons can be burst by 2 arrows:
- Shoot an arrow at x = 2, bursting the balloons [1,2] and [2,3].
- Shoot an arrow at x = 4, bursting the balloons [3,4] and [4,5].
Constraints:
• `1 <= points.length <= 10^5`
• `points[i].length == 2`
• `-2^31 <= xstart < xend <= 2^31 - 1`

### 解題

func findMinArrowShots(points [][]int) int {
sort.Slice(points, func(i, j int) bool {
return points[i] < points[j] // 根據氣球的右邊最排序，待會從左邊開始戳破
})
arrowPos := points
ans := 1
for i:=1; i<len(points); i++ {
if (arrowPos >= points[i]) { //戳不到
continue
}
ans++ //多加一根針
arrowPos = points[i] //戳當下氣球的最後面
}
return ans
}