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:
容器装水量 = 两点横坐标差值 * 两点纵坐标的最小值
要不重不漏的找出可以组成容器的所有组合：定义两个指针，i，j分别指向数组的前后端，计算容器装水量，取result较大值。并移动一步指针（either i or j, 两个指针不能同时移动），计算面积取较大值，直到i , j相遇。
注意：移动i还是j，看他们的height谁小就移动谁
j初始值是数组长度减1 ，不然再循环中访问不到数组元素，会抛出outofboundary异常。
代码：
class Solution { public int maxArea(int[] height) { int res = 0, i = 0, j = height.length - 1; while(i < j) { res = Math.max(res, Math.min(height[i], height[j]) * (j - i)); if (height[i] < height[j]) i++; else j--; } return res; } }