動機

題目給的是氣球的寬度

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: 2Explanation: 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

 

Constraints:

  • 1 <= points.length <= 104
  • points[i].length == 2
  • -231 <= xstart < xend <= 231 - 1

Sol

如果有一個範圍完整包含另一個，就只要打小的另一個也會被打到

如果是只有交疊一部份，就是只留交疊的部分

剩下就只能另外打了

class Solution:
    def findMinArrowShots(self, ps: List[List[int]]) -> int:
        stk = []
        
        ps.sort(key=lambda x: (x[0], x[1]-x[0]))
        #print(ps)
        stk.append(ps[0])
        for (a,b) in ps[1:]:
            x,y = stk[-1]
            if x <= a <= y and x <= b <= y:
                stk.pop()
                stk.append([a,b])
            elif x <= a <= y:
                stk.pop()
                stk.append([a,y])
            else:
                stk.append([a,b])
        return len(stk)