Leetcode 452. Minimum Number of Arrows to Burst Balloons

Description: 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.

Link: 452. Minimum Number of Arrows to Burst Balloons

Examples:

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

Example 4:
Input: points = [[1,2]]
Output: 1

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

思路: 至少需要多少只箭射穿所有气球，只要xstart ≤ x ≤ xend. 这个题会让人联想到大小信封套进去的哪个题目。如果两个区间有交集，就可以用同一只箭射穿，所以按照区间的尾排序，如果另一个的开头小于等于尾，那么有交集。

if len(points) == 0: return 0
        points.sort(key=lambda tup: tup[1])
        cur, res = points[0][1], 1
        for i in range(len(points)):
            if points[i][0] <= cur:
                continue
            cur = points[i][1]
            res += 1
        return res

日期: 2021-04-02