# 【LeetCode】 11. 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.

> 給予n個非負整數a1~an，代表一個座標點(i,ai)。
> 現在我們畫了n條從(i,ai)到(i,0)的直線。
> 將兩條線和x軸當作一個容器，請找出哪兩條的容量是最大的。
> 提示：你不能傾斜容器，且n最小是2。

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

## Example:

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

## Solution

* 參考至[這裡](https://leetcode.com/problems/container-with-most-water/discuss/6126/C++-O(n)-solution-with-thought-process-applying-simple-bucket-theory)。
* 從左邊和右邊往中間逼近，左和右誰比較矮，誰就要往中間動一步。
* 每動一步就計算一次容量，記憶最大值。

### Code

class Solution {
public:
    int maxArea(vector<int>& height) {
        int ans=0;
        for(int i = 0,j=(int)height.size()-1;i!=j;height[i]>height[j]?j--:i++)
        {
            int temp = min(height[i],height[j])*(j-i);
            ans = ans>temp?ans:temp;
        }
        return ans;
    }
};

###### tags: `LeetCode` `C++`