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 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

这个题想了很久，结果没想到怎么做，其实现在理解以下就是一个双边贪心的策略，尽量的宽，又尽力的高才能存储更多的水，所以，从两端向中间贪心，如果左边的比右边的高就更新右边的边，往中间靠一阶，看能不能得到更好的，同理，如果右边的比左边的短，右边就向右扫描一个，如果相等，那随便选择一边更新一格就可以。

 class Solution {

 public:

     int maxArea(vector<int>& height) {

         int r = ; int l = height.size()-;

         int Max = (min(height[l],height[r])*(l-r));

         while(r<l){

             if(height[r] <= height[l]) r++;

             else if(height[l] < height[r]) l--;

             Max = max(Max,(min(height[l],height[r])*(l-r)));

         }

         return Max;

     }

 };