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# LeetCode 42. Trapping Rain Water

Given `n` non-negative integers representing an elevation map where the width of each bar is `1`, compute how much water it can trap after raining.

**Example 1:**![](https://assets.leetcode.com/uploads/2018/10/22/rainwatertrap.png)

```
Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.
```

**Example 2:**

```
Input: height = [4,2,0,3,2,5]
Output: 9
```

**Constraints:**

* `n == height.length`
* `0 <= n <= 3 * 10^4`
* `0 <= height[i] <= 10^5`

## Solution

[English Version in Youtube](https://youtu.be/1LXscFcXkOk)

[中文版解答Youtube Link](https://youtu.be/vzNQr6ocvt4)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1BK4y1N75u/)

```
class Solution {
public:
    int trap(vector<int>& height) {
        int i = 0;
        int j = height.size() - 1;
        int maxleft = 0, maxright = 0;
        int res = 0;
        while(i < j) {
            if(height[i] <= height[j]) {
                if (height[i] > maxleft) maxleft = height[i];
                else res += (maxleft - height[i]);
                i++;
            } else {
                if (height[j] > maxright) maxright = height[j];
                else res += (maxright - height[j]);
                j--;
            }
        }
        return res;
    }
};
```