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class Solution:
def maxArea(self, height) -> int:
# 设x,y是某两根柱子的索引
# 要找的是表达式 abs(x-y) * min(height[x],height[y]) 之最大值
# 初始化两个指针，分别指向数组头和尾的索引
# 若向内移动长板,min()函数的值不变或变小，s必然变小
# 若向内移动短板,min()函数的值可能变大，s可能变大
i,j = 0,len(height) - 1
cur_max = (j-i) * min(height[i],height[j])
while i<j:
dis = j - i
cur_v = dis * min(height[i],height[j])
if cur_v > cur_max:
cur_max = cur_v
else:
pass
if height[i] <= height[j]:
i += 1
j -= 1
return cur_max
