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.

思路：从两侧往中间遍历。使用curHeight保存当前计算过的容器的最大高度。由于遍历是从两侧往中间进行的，因此，这将保证任何高度不大于curHeight的容器的最大容量已经被计算过。于是，我们只需在当前两端的高度都大于curHeight时，才更新curHeight，同时计算以新高度为高度的容器的最大容量。

 1 class Solution { 2 public: 3     int maxArea( vector<int> &height ) { 4         int area = 0, curHeight = 0; 5         int left = 0, right = (int)height.size()-1; 6         while( left < right ) { 7             if( height[left] <= curHeight ) { ++left; continue; } 8             if( height[right] <= curHeight ) { --right; continue; } 9             curHeight = min( height[left], height[right] );10             area = max( area, (right-left)*curHeight );11         }12         return area;13     }14 };

Container With Most Water