# Container With Most Water

Given n non-negative integers a1a2, ..., an, where each represents a point at coordinate (iai). n vertical lines are drawn such that the two endpoints of line i is at (iai) 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.

C++实现如下：

#include<iostream>
#include<vector>
using namespace std;

class Solution
{
public:
int maxArea(vector<int> &height)
{
if(height.empty()||height.size()==1)
return 0;
int n=height.size();
int i,j;
int maxArea=0;
for(i=0; i<n-1; i++)
{
for(j=i+1; j<n; j++)
{
int tmp=min(height[i],height[j])*(j-i);
if(maxArea<tmp)
maxArea=tmp;
}
}
return maxArea;
}
};

int main()
{
Solution s;
vector<int> vec={2,4,1,3,0,6};
cout<<s.maxArea(vec)<<endl;
}

O(n)的复杂度。保持两个指针i,j；分别指向长度数组的首尾。如果ai 小于aj，则移动i向后（i++）。反之，移动j向前（j--）。如果当前的area大于了所记录的area，替换之。这个想法的基础是，如果i的长度小于j，无论如何移动j，短板在i，不可能找到比当前记录的area更大的值了，只能通过移动i来找到新的可能的更大面积。

C++实现代码如下：

#include<iostream>
#include<vector>
using namespace std;

class Solution
{
public:
int maxArea(vector<int> &height)
{
if(height.empty()||height.size()==1)
return 0;
int n=height.size();
int i,j;
int maxArea=0;
i=0;
j=n-1;
while(i<j)
{
int tmp=min(height[i],height[j])*(j-i);
if(maxArea<tmp)
maxArea=tmp;
if(height[i]<height[j])
i++;
else
j--;
}
return maxArea;
}
};

int main()
{
Solution s;
vector<int> vec= {2,4,1,3,0,6};
cout<<s.maxArea(vec)<<endl;
}

posted @ 2014-11-19 20:39  Jessica程序猿  阅读(187)  评论(0编辑  收藏  举报