题目：
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.
题解：
这道题挺类似二分查找法的，这道题是先从两头开始算面积，面积的计算要由短板决定，并维护一个当前最大面积。
然后逐步替换小的短板来计算面积。每一步只替换短板的原因是，短板决定面积，而高板不影响，所以要想有所改变就改变那个有决定性的东西。。
网上有好多人都画出图来了。。可以搜搜看。。。
代码如下：
1 public int maxArea(int[] height) {
2 if(height == null || height.length == 0)
3 return 0;
4
5 int low = 0, high = height.length -1 ;
6 int max = 0;
7 while (low < high) {
8 int area = (high-low)*Math.min(height[low], height[high]);
9
10 max = Math.max(max, area);
11 if (height[low] < height[high])
12 low++;
13 else
14 high--;
15 }
16 return max;
17 }
2 if(height == null || height.length == 0)
3 return 0;
4
5 int low = 0, high = height.length -1 ;
6 int max = 0;
7 while (low < high) {
8 int area = (high-low)*Math.min(height[low], height[high]);
9
10 max = Math.max(max, area);
11 if (height[low] < height[high])
12 low++;
13 else
14 high--;
15 }
16 return max;
17 }