problem：

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.

thinking:

(1)短边决定水箱的有效高，底要尽可能的宽。

（2）典型的双指针求解的题型。

（3）贪心的策略，哪条边短，往里收缩寻找下一条边。

code:

class Solution {
public:
    int maxArea(vector<int> &height) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int i = 0;
        int j = height.size() - 1;
        
        int ret = 0;
        while(i < j)
        {
            int area = (j - i) * min(height[i], height[j]);
            ret = max(ret, area);
            
            if (height[i] <= height[j])
                i++;
            else
                j--;
        }
        
        return ret;
    }
};

时间复杂度为O（n）

int area(vector<int>::iterator &a,vector<int>::iterator &b)
{
    return (b-a)*(*a>*b?*b:*a);

}

class Solution {
public:
    int maxArea(vector<int> &height) {
        int max_area=0;
        for(vector<int>::iterator i=height.begin()+1;i!=height.end();i++)
        {
            for(vector<int>::iterator j=height.begin();j!=i;j++)
            {
                max_area=max(max_area,area(j,i));
            }
        }
        return max_area;


    }
};