> For the complete documentation index, see [llms.txt](https://hao-fu-1.gitbook.io/oj/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hao-fu-1.gitbook.io/oj/multiple-pointers/two_sum/container-with-most-water.md).

# Container With Most Water

<https://leetcode.com/problems/container-with-most-water/description/>

> 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 and n is at least 2.

## Thoughts

\[i, j]看作是当前问题的范围，假设i是短边，暴力解f\_i(i+1, j)是对i遍历所有\[i + 1, j],  因此f(i, j) = max(f\_i(i+1, j), f(i + 1, j), fij),  但因为高度受限于i, f\_i(i + 1, j)是一定小于fij的，因此可省去对i遍历所有\[i + 1, j]，问题范围缩小至\[i + 1, j]。

## Code

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

## Analysis

时间复杂度O(n). 空间O(1).