動機

一開始以為要用stack，因為中間不用的柱子可以忽略掉，但是這樣沒辦法保留距離短但比較遠的柱子。

Problem

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 the line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.

Notice that you may not slant the container.

 

Example 1:

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

Input: height = [1,1]Output: 1

Example 3:

Input: height = [4,3,2,1,4]Output: 16

Example 4:

Input: height = [1,2,1]Output: 2

 

Constraints:

  • n == height.length
  • 2 <= n <= 105
  • 0 <= height[i] <= 104

Sol

之後看著圖，想到範圍就想到，從最外層往裡面縮，看起來可以用two ptr

但用two ptr之前要先能確定，怎麼變大與縮小，而我們有柱子的長度!! 把低的丟掉就好了

class Solution:
    def maxArea(self, hs: List[int]) -> int:
        a, b, ret = 0, len(hs)-1, 0
        while a < b:
            water = (b-a) * min(hs[a],hs[b])
            ret = max(ret, water)
            if hs[a] >= hs[b]:
                b -= 1
            else:
                a += 1
        return ret