<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) {
       int height_size = height.size();
       int left_max = height[0];
       int right_max = height[height_size-1];
       int left = 0;
       int right = height_size-1;
       int result = 0;
       while(left < right)
       {
            int temp = 0;
            if(left_max > right_max)
            {
                temp = height[right-1];
                if(temp > right_max)
                    right_max = temp;
                else
                    result += right_max - temp;
                right--;
            }
            else
            {
                temp = height[left+1];
                if(temp > left_max)
                    left_max = temp;
                else
                    result += left_max - temp;
                left++;
            }
       }
       return result;
    }
};
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