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.

思路：

思路1为两层遍历，思路2类似
http://blog.csdn.net/lanxu_yy/article/details/17335239的思路。利用一个数组保存可能的水池左侧，另一个数组保存可能的水池右侧。

代码：

思路1：

class Solution {
public:
    int maxArea(vector<int> &height) {
        int max=0;
        for(int i=0;i<height.size();i++)
        {
            for(int j=i+1;j<height.size();j++)
            {
                int a = area(height,i,j);
                if(a>max)
                    max=a;
            }
        }
        return max;
    }
    
    int area(vector<int> &height, int i, int j)
    {
        int h = height[i]<height[j]?height[i]:height[j];
        return h*(j-i);
    }
};

思路2：

class Solution {
public:
    vector<int> begin;
    vector<int> ibegin;
    vector<int> end;
    vector<int> iend;
    int maxArea(vector<int> &height) {
        if(height.size()<=1)
            return 0;
        begin.push_back(height[0]);
        ibegin.push_back(0);
        end.push_back(height[1]);
        iend.push_back(1);
        
        for(int i=1;i<height.size();i++)
        {
            if(begin[begin.size()-1]<height[i])
            {
                begin.push_back(height[i]);
                ibegin.push_back(i);
            }
                
            if(end[end.size()-1]<=height[i])
            {
                end[end.size()-1] = height[i];
                iend[end.size()-1] = i;
            }
            else
            {
                end.push_back(height[i]);
                iend.push_back(i);
            }
        }
        
        int max = 0;
        
        for(int i=0;i<begin.size();i++)
        {
            for(int j=0;j<end.size();j++)
            {
                if(ibegin[i]<iend[j])
                {
                    int area = (begin[i]<end[j]?begin[i]:end[j])*(iend[j]-ibegin[i]);
                    if(area > max)
                        max = area;
                }
            }
        }
        return max;
    }
};