本文主要是介绍[LeetCode] 11.Container With Most Water，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目内容
给定 n 个非负整数 a1，a2，...，an，每个数代表坐标中的一个点 (i, ai) 。在坐标内画 n 条垂直线，垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0)。找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/container-with-most-water
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
https://leetcode-cn.com/problems/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题目思路
可以采用从两边到中间收的方法，每一次计算后较低的那一侧向另一侧靠近，知道两者的间隔只剩下1为止。
程序代码
class Solution(object):def maxArea(self, height):""":type height: List[int]:rtype: int"""max_area=0left,right=0,len(height)-1while left<right:area=min(height[left],height[right])*(right-left)max_area=max(area,max_area)if height[left]<=height[right]:temp=height[left]while left<right and height[left]<=temp:left+=1else:temp=height[right]while left<right and height[right]<=temp:right-=1return max_area
这篇关于[LeetCode] 11.Container With Most Water的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
