资讯专栏INFORMATION COLUMN

容器最大盛水量

luckyw / 728人阅读

摘要:容器最大盛水量给定个非负整数,,,,其中每个表示坐标,处的点。找到两条线,它们与轴一起形成一个容器,使得容器含有最多的水。

容器最大盛水量 Container With Most Water

给定n个非负整数a1,a2,...,an,其中每个表示坐标(i,ai)处的点。

绘制n条垂直线,使得线i的两个端点在(i,ai)和(i,0)处。

找到两条线,它们与x轴一起形成一个容器,使得容器含有最多的水。

注意:您不得倾斜容器,n至少为2。

Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai).

n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0).

Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

example 1

input: [1,2,4,3]
output: 4

note: arr[1] 和 arr[3]的间隔(width)为2,min(arr[1], arr[3]) = 2,
max_water = 2 * 2 = 4 

example 2

input: [2,3,10,5,7,8,9]
output: 36

example 3

input: 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output: 4913370
思路

考虑到效率问题,两层for循环两两组合数组中的元素为最基本方法,时间复杂度为O(n²),在example3中明显会导致运行时间太长

由于容器大小和两个因素有关,width和height,假设初始最大盛水量的状态为width = len(arr) - 1,height = min(arr[0], arr[len[arr] - 1])

head = 0, tail = len(arr) - 1, 采用头尾两个下标往中间移动靠拢,扫描整个数组,head移动还是tail移动取决于arr[head]arr[tail]谁比较小(容器最大盛水量取决于短板),每移动一次计算max_water是否增大

由于只扫描了一次数组,时间复杂度为O(n)

代码
class Solution(object):

    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int 返回值为最大盛水量
        """
        head = 0
        tail = len(height) - 1
        max_water = 0
        while head < tail:
            max_water = max(max_water, min(height[head], height[tail]) * (tail - head))
            if height[head] < height[tail]:
                head += 1
            else:
                tail -= 1
        return max_water

本题以及其它leetcode题目代码github地址: github地址

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