Leetcode 11 Container with most Water

Leetcode 11 Container with most Water"/>

Leetcode 11 Container with most Water

Leetcode 11 Container with most Water

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.

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

来源：力扣（LeetCode）
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思路解析

• 思路一：暴力穷举法

最简单的思路就是穷举所有的组合，求出最大面积

class Solution:def maxArea(self, height: List[int]) -> int:if height == [] or len(height) == 1 or (len(height) == 2 and 0 in height):return 0max = 0new_height = enumerate(height)for index, value in new_height:next_index = index + 1while next_index < len(height):area = (next_index - index) * min(height[next_index], value)if max < area:max = areanext_index = next_index + 1return max            

但这种解法的缺点也显而易见，时间复杂度过高，为 O ( n 2 ) O(n^2) O(n2)

• 思路二

利用双指针，首指针从数组首部开始移动，尾指针从尾部开始移动，只遍历可能比当前面积大的组合，而且时间复杂度为 O ( n ) O(n) O(n)

怎么遍历比当前遍历可能比当前面积大的组合呢？

向指向长指针的方向移动短指针。这样做的影响是容器宽度减一，但高度可能增大，也就是面积可能增大。

而如果向指向短指针的方向移动长指针的话，容器宽度减一，高度减小或者依然为短指针所在位置的高度，导致面积减小或者不变。

所以我们采用向指向长指针的方向移动短指针，使得我们可以在更小的组合集合中找到使面积最大的组合。

class Solution:def maxArea(self, height: List[int]) -> int:length = len(height)left = 0right = length - 1max_area = 0while left < right:area = min(height[left], height[right]) * (right - left)max_area = max_area if area < max_area else areaif height[left] < height[right]:left = left + 1else:right = right - 1return max_area

双指针法正确性的证明