題目
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.

答案
Idea:
start with l = 0, r = height.length - 1

record the max container area between two lines, height[l] and height[r]
compare height[l] and height[r], if one is shorter, for example, height[l] < height[r], then moving l to the right may find a taller line such that the resulting container holds more water.

The explanation above is more intuitive than it is rigorous.
Here let me try to explain why it should work:

Given a height array of size n
left line is at index 0
right line is at index n - 1
if the left line is shorter than the right line, then 'left++' gets you a possible candidate that may(or may not result larger container area). but with 'right--', you are guaranteed no larger container area(because container area is determined by the shorter line).

Now, we consider a smaller range of the height array(from index 1 to n -1), the logic still applies. Keep recursing until the left >= right, then we will stop searching.

class Solution {
    public int maxArea(int[] height) {
        if(height.length == 0) return 0;
        int l = 0, r = height.length - 1;
        int max = 0;
        while(l < r) {
            int w = r - l;
            int h = Math.min(height[l], height[r]);
            max = Math.max(max, w * h);
            
            if(height[l] < height[r]) {
                l++;
            }
            else {
                r--;
            }
        }
        return max;
    }
}