#include <iostream>

 #include <vector>

 #include <algorithm>

 using namespace std;

 class Solution {

 public:

     int maxArea(vector<int> &height) {

         if (height.empty()) return ;

         int i = , j = height.size() - ;

         int max = INT_MIN;

         int area;

         while (i < j)

         {

             area = min(height[i],height[j])*(j-i);

             if (area>max) max = area;

             if (height[i] < height[j]) i++;//谁是短板，谁就该移动

             else j--;

         }

         return max;

     }

 };

 int main()

 {

     Solution s;

     int array[] = {,,,,};

     vector<int> h(array,array+);

     cout << s.maxArea(h)<<endl;

     return ;

 }

看了nandawys的评论，找到了O(n)方法，思路是从两头到中间扫描，设i,j分别指向height数组的首尾。
那么当前的area是min(height[i],height[j]) * (j-i)。
当height[i] < height[j]的时候，我们把i往后移，否则把j往前移，直到两者相遇。

这个正确性如何证明呢？
代码里面的注释说得比较清楚了，即每一步操作都能保证当前位置能取得的最大面积已经记录过了，而最开始初始化的时候最大面积记录过，所以有点类似于数学归纳法，证明这个算法是正确的。

Container With Most Water
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.

代码
600ms过大集合

class Solution {

public:

    int maxArea(vector<int> &height) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        int max = 0;

        for( int i = 0 ; i < height.size(); ++i)

        {

            int hi = height[i];

            if(hi == 0)

                continue;

            int minPosibleIndex = max / hi + i;

            for(int j = height.size() - 1; j > i && j >= minPosibleIndex; --j)

            {

                int hj = height[j];

                int area = min(hi,hj) * (j - i);

                if (area > max)

                    max = area;

            }

        }

        return max;

    }

};

Code rewrite at 2013-1-4,O(n)

 class Solution {

 public:

     int maxArea(vector<int> &height) {

         if (height.size() < ) return ;

         int i = , j = height.size() - ;

         int maxarea = ;

         while(i < j) {

             int area = ;

             if(height[i] < height[j]) {

                 area = height[i] * (j-i);

                 //Since i is lower than j,

                 //so there will be no jj < j that make the area from i,jj

                 //is greater than area from i,j

                 //so the maximum area that can benefit from i is already recorded.

                 //thus, we move i forward.

                 //因为i是短板，所以如果无论j往前移动到什么位置，都不可能产生比area更大的面积

                 //换句话所，i能形成的最大面积已经找到了，所以可以将i向前移。

                 ++i;

             } else {

                 area = height[j] * (j-i);

                 //the same reason as above

                 //同理

                 --j;

             }

             if(maxarea < area) maxarea = area;

         }

         return maxarea;

     }

 };