［抄題］：

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 xendbursts 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).

[暴力解法]：

時間分析：

空間分析：

[優化後]：

時間分析：

空間分析：

［奇葩輸出條件］：

［奇葩corner case］：

［思維問題］：

知道是掃描線，忘了怎麽寫了：更新結尾。必要時+count

［英文數據結構或算法，為什麽不用別的數據結構或算法］：

掃描線要先對取件進行排序。

［一句話思路］：

［輸入量］：空： 正常情況：特大：特小：程序裏處理到的特殊情況：異常情況（不合法不合理的輸入）：

［畫圖］：

［一刷］：

1. 循環過程中要依據end來進行更新

［二刷］：

［三刷］：

［四刷］：

［五刷］：

[五分鐘肉眼debug的結果]：

［總結］：

［復雜度］：Time complexity: O() Space complexity: O()

［算法思想：叠代/遞歸/分治/貪心］：

［關鍵模板化代碼］：

［其他解法］：

［Follow Up］：

［LC給出的題目變變變］：

[代碼風格] ：

[是否頭一次寫此類driver funcion的代碼] ：

[潛臺詞] ：

452. Minimum Number of Arrows to Burst Balloons紮氣球的個數最少