1 #include <iostream> 2 #include <vector> 3 #include <algorithm> 4 using namespace std; 5  6 class Solution { 7 public: 8     int maxArea(vector<int> &height) { 9         if (height.empty()) return 0;10         int i = 0, j = height.size() - 1;11         int max = INT_MIN;12         int area;13         while (i < j)14         {15             area = min(height[i],height[j])*(j-i);16             if (area>max) max = area;17             if (height[i] < height[j]) i++;//谁是短板，谁就该移动18             else j--;19         }20         return max;21     }22 };23 24 int main()25 {26     Solution s;27     int array[] = {2,5,3,1,6};28     vector<int> h(array,array+5);29     cout << s.maxArea(h)<<endl;30     return 0;31 }

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)

 1 class Solution { 2 public: 3     int maxArea(vector<int> &height) { 4         if (height.size() < 2) return 0; 5         int i = 0, j = height.size() - 1; 6         int maxarea = 0; 7         while(i < j) { 8             int area = 0; 9             if(height[i] < height[j]) {10                 area = height[i] * (j-i);11                 //Since i is lower than j, 12                 //so there will be no jj < j that make the area from i,jj 13                 //is greater than area from i,j14                 //so the maximum area that can benefit from i is already recorded.15                 //thus, we move i forward.16                 //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17                 //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。18                 ++i;19             } else {20                 area = height[j] * (j-i);21                 //the same reason as above22                 //同理23                 --j;24             }25             if(maxarea < area) maxarea = area;26         }27         return maxarea;28     }29 };