452. Minimum Number of Arrows to

2018-03-09  本文已影响0人  Jonddy
题目要求:

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 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 travelling up infinitely. The problem is to find the minimum number of arrows that must be shot to burst all balloons.

Examples:

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).

解题思路:
代码:

class Solution(object):
    def findMinArrowShots(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        if not points:
            return False

        points = sorted(points, key=lambda x: x[1])

        curmax, num = points[0][1], 1
        for balloon in points:
            if balloon[0] > curmax:
                curmax = balloon[1]
                num += 1
        return num

if __name__ == "__main__":
    print(Solution.findMinArrowShots(self=None, points = [[9,12],[1,10],[4,11],[8,12],[3,9],[6,9],[6,7]]))
python:sort()函数和lambda表达式

iterable:是可迭代类型;
cmp:用于比较的函数,比较什么由key决定,有默认值,迭代集合中的一项;
key:用列表元素的某个属性和函数进行作为关键字,有默认值,迭代集合中的一项;
reverse:排序规则. reverse = True 或者 reverse = False,有默认值。
返回值:是一个经过排序的可迭代类型,与iterable一样。
key和cmp可以实现相同的功能,key的效率高于cmp。

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