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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://assets.leetcode.com/uploads/2018/10/22/rainwatertrap.png" style="width: 412px; height: 161px;" />
<pre>
<strong>Input:</strong> height = [0,1,0,2,1,0,1,3,2,1,2,1]
<strong>Output:</strong> 6
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> height = [4,2,0,3,2,5]
<strong>Output:</strong> 9
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == height.length</code></li>
	<li><code>1 &lt;= n &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= height[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    int trap(vector<int>& height) {
        vector<int> maxl(height.size()), maxr(height.size());
        int v = height[0];

        maxl.push_back(0);
        for(int i = 1; i < height.size(); i++) {
            maxl[i] = v;
            //cout << maxl[i] << " ";
            if(height[i] > v) v = height[i];
        }
        //cout << endl;
        maxr[height.size()-1] = 0;
        v = height[height.size()-1];
        for(int i = height.size()-2; i >= 0; i--) {
            maxr[i] = v;
            //cout << maxr[i] << " ";
            if(height[i] > v) v = height[i];
        }

        //cout << endl;
        int res = 0;
        for(int i = 0; i < height.size(); i++) {
            v = min(maxl[i], maxr[i]) - height[i];
            //cout << v << " ";
            if(v > 0) res += v;
        }

        return res;
    }
};
24000