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
思路： 最左与最右分别设置两个点：l,r, 设它们的高为：hl,hr 先记录一次面积: area = (r-l)*min(hl,hr) 即为：两点间的距离*两者之间最矮的高度 假设在l,r原始点之内仍有可能存在更大的面积组合: 因为要往里找新的l,r组合，长度（r-l)是一直在缩小的，如果存在更大面积，则一定存在更长的高，则min(*hr,*hl)>min(hr,hl) 假设一开始，hr > hl, area = (r-l) * hl ;如果存在更大面积， 则*hl > hr, *area = (r-*l) * hr，因hr > hl, *area有可能会大于原面积 假设，hr < hl, area = (r-l) * hr ;如果存在更大面积， 则*hr > hl, *area = (*r-l) * hl, 因hl > hr, *area有可能会大于原面积 如果hl == hr 怎么办？ 不断同时移动两个点，直到新的lr的高度同时大于原来的高度： 因为r-l一直在缩小，min(hl,hr)必须变大，所以hr,hl要同时变大才，有机会超越原来的面积。
class Solution(object): def maxArea(self, height): """ :type height: List[int] :rtype: int """ l,r = 0, len(height)-1 box = [self.getarea(height, l, r)] while l < r: if l < r and height[l] < height[r]: l += 1 if l < r and height[l] > height[l-1]: box.append(self.getarea(height, l, r)) if l < r and height[l] > height[r]: r -= 1 if l < r and height[r] > height[r+1]: box.append(self.getarea(height, l, r)) if l < r and height[l] == height[r]: now = height[l] while l < r and now >= height[l]: l += 1 while l < r and now >= height[r]: r -= 1 box.append(self.getarea(height, l, r)) return max(box) def getarea(self, height, l, r): return (r-l)*min(height[l], height[r])