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資訊專欄INFORMATION COLUMN

容器最大盛水量

luckyw / 731人閱讀

摘要:容器最大盛水量給定個非負整數,,,,其中每個表示坐標,處的點。找到兩條線,它們與軸一起形成一個容器,使得容器含有最多的水。

容器最大盛水量 Container With Most Water

給定n個非負整數a1,a2,...,an,其中每個表示坐標(i,ai)處的點。

繪制n條垂直線,使得線i的兩個端點在(i,ai)和(i,0)處。

找到兩條線,它們與x軸一起形成一個容器,使得容器含有最多的水。

注意:您不得傾斜容器,n至少為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]的間隔(width)為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

考慮到效率問題,兩層for循環兩兩組合數組中的元素為最基本方法,時間復雜度為O(n2),在example3中明顯會導致運行時間太長

由于容器大小和兩個因素有關,width和height,假設初始最大盛水量的狀態為width = len(arr) - 1,height = min(arr[0], arr[len[arr] - 1])

head = 0, tail = len(arr) - 1, 采用頭尾兩個下標往中間移動靠攏,掃描整個數組,head移動還是tail移動取決于arr[head]arr[tail]誰比較小(容器最大盛水量取決于短板),每移動一次計算max_water是否增大

由于只掃描了一次數組,時間復雜度為O(n)

代碼
class Solution(object):

    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int 返回值為最大盛水量
        """
        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

本題以及其它leetcode題目代碼github地址: github地址

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