用最少的弓箭射穿所有氣球 Min of Arrow to Burst Balloon_Leetcode #452精選75

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題目敘述

題目會給定一個輸入陣列points，陣列元素都是一組pair，

points[i] = [start_i, end_i]，分別代表每顆氣球的左邊界，和右邊界。

假設弓箭射出去後，動能不會減弱，沿路上的氣球都會被射穿。

請問最少需要幾隻弓箭，才可以射穿所有氣球?

題目的原文敘述

測試範例

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].

約束條件

Constraints:

• 1 <= points.length <= 10^5

輸入陣列長度介於1~十萬之間。

• points[i].length == 2

每個陣列元素都是一組Pair

• -2^31 <= xstart < xend <= 2^31 - 1

每顆氣球的左邊界、右邊界介於 32bit整數範圍之內。

演算法 Greedy + 排序

和前一題 排序應用題: 不重複的區間 有異曲同工之妙。

想要讓弓箭的使用數量最精簡，就是用同一支箭去射穿區間有重複的氣球。

可以想到先讓所有區間都依終點的大小作排序。