[leetcode]Container With Most Water(using Python)
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题目描述:
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
思路:最容易想到的方法是嵌套循环,把所有的container的容量都求出来,最后得到最大的容量,但是很显然这样的做法的时间复杂度是O(n^2),会超出运行时间限制。进一步思考发现,container的容量是由底长和较短的那个壁决定的,我们使用两个指针 left 和 right,一个在List[0],另一个在List[len(List) - 1],而一旦能够确定当前容器的左边或者右边的壁较短,那么由这个较短壁能够构成的容器的最大容量就已经求得了。之后需要将这个较短壁的指针向后或向前移动一位(具体看是哪个指针指向当前较短壁,若是left就后移一位,若是right就向前移一位)。Python代码如下:
class Solution(object): def maxArea(self, height): """ :type height: List[int] :rtype: int """ if len(height) < 2: return -1 else: mwater = 0 left = 0 right = len(height) - 1 while left < right: mwater = max(mwater, (right - left)*min(height[left],height[right])) if height[left] < height[right]: left +=1 else: right -= 1 return mwater
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