# 盛最多水的容器
给你 n 个非负整数 a1，a2，...，an，每个数代表坐标中的一个点 (i, ai) 。在坐标内画 n 条垂直线，垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0) 。找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。
说明：你不能倾斜容器。
示例 1：
输入：[1,8,6,2,5,4,8,3,7]
输出：49
解释：图中垂直线代表输入数组 [1,8,6,2,5,4,8,3,7]。在此情况下，容器能够容纳水（表示为蓝色部分）的最大值为 49。
示例 2：
输入：height = [1,1]
输出：1
示例 3：
输入：height = [4,3,2,1,4]
输出：16
示例 4：
输入：height = [1,2,1]
输出：2
提示：
n = height.length2 <= n <= 3 * 1040 <= height[i] <= 3 * 104
以下错误的选项是？
## aop
### before
```cpp
#include
using namespace std;
```
### after
```cpp
int main()
{
Solution sol;
int arr1[] = {1, 8, 6, 2, 5, 4, 8, 3, 7};
int length1 = sizeof(arr1) / sizeof(arr1[0]);
vector nums1(arr1, arr1 + length1);
cout << sol.maxArea(nums1) << endl;
return 0;
}
```
## 答案
```cpp
class Solution
{
public:
int maxArea(vector &height)
{
int res = 0;
for (int i = 0; i < height.size(); i++)
{
for (int j = i + 1; j < height.size(); j++)
{
int temp1 = max(height[i], height[j]);
int temp2 = temp1 * (j - i);
res = min(res, temp2);
}
}
return res;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
int maxArea(vector &height)
{
int i = 0;
int j = height.size() - 1;
int res = 0;
while (i < j)
{
int temp1 = min(height[i], height[j]);
int temp2 = temp1 * (j - i);
res = max(res, temp2);
if (height[i] < height[j])
i++;
else
j--;
}
return res;
}
};
```
### B
```cpp
class Solution
{
public:
int maxArea(vector &height)
{
int tmp = 0, res = 0, max = 0;
for (int i = 0; i < height.size(); ++i)
{
for (int j = i + 1; j < height.size(); ++j)
{
if (height[i] < height[j])
{
tmp = height[i];
}
else
{
tmp = height[j];
}
res = tmp * (j - i);
if (res > max)
{
max = res;
}
}
}
return max;
}
};
```
### C
```cpp
class Solution
{
public:
int maxArea(vector &height)
{
int tail = height.size() - 1;
int head = 0;
int maxarea = height.at(head) > height.at(tail) ? height.at(tail) * (tail - head) : height.at(head) * (tail - head);
for (int i = 0; i < height.size(); i++)
{
if (maxarea < (height.at(head) > height.at(tail) ? height.at(tail) * (tail - head) : height.at(head) * (tail - head)))
maxarea = height.at(head) > height.at(tail) ? height.at(tail) * (tail - head) : height.at(head) * (tail - head);
if (height.at(head) > height.at(tail))
{
tail--;
}
else
{
head++;
}
}
return maxarea;
}
};
```