Given n non-negative integers a₁, a₂, ..., a_n , where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) 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

思路

1. 暴力遍历，依次遍历容器的左右边界，以左右边界中高度较小的当做容器高度，计算面积。

# 1. 使用暴力搜索，时间复杂度为O(n**2)
    def maxArea(self, height: List[int]) -> int:
        max_area = 0
        for i in range(0, len(height) - 1):
            for j in range(i + 1, len(height)):
                area = (j - i) * min(height[i], height[j])
                max_area = max(area, max_area)
        return max_area

2. 双指针法， 先以数组的最左最右当做容器的左右边界，再向中间收敛。因为从数组两端开始已经保证了容器的最宽宽度，现在要找更高的高度，如果左边界高度小于右边界，则左边界向里移动，否则右边界向内移动。

# 2. 双指针，向内收敛，时间复杂度为O(n)
    def maxArea2(self, height: List[int]) -> int:
        l, r = 0, len(height) - 1
        max_area = 0
        while (l < r):
            area = (r - l) * min(height[l], height[r])
            max_area = max(max_area, area)
            if height[l] < height[r]:
                l += 1
            else:
                r -= 1

        return max_area