Leetcode Container With Most Water

丹

发布于 2018-09-04 11:37:02

4750

发布于 2018-09-04 11:37:02

题目：Container With Most Water

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.

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

解答：

第一想法两个for循环，但是很明显没有通过。

参考：[LeetCode]题解（python）：011-Container With Most Water

class Solution(object):
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        size = len(height) # the size of height
        maxm = 0 # record the most water
        j = 0
        k = size - 1
        while j < k:
            if height[j] <= height[k]:
                maxm = max(maxm,height[j] * (k - j))
                j += 1
            else:
                maxm = max(maxm,height[k] * (k - j))
                k -= 1
        return maxm

效率更高的方法：sample 48 ms submission

class Solution:
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """        
        i,j,max_v=0,len(height)-1,0
        while i<j:
            if height[i]<height[j]:
                short_p=height[i]
                v=short_p*(j-i)
                i+=1
                while i<j and height[i]<=short_p:
                    i+=1
            else:
                short_p=height[j]
                v=short_p*(j-i)
                j-=1
                while i<j and height[j]<=short_p:
                    j-=1
            if v>max_v:
                max_v=v
        return max_v

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原始发表：2018年08月14日，如有侵权请联系 cloudcommunity@tencent.com 删除

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