leetcode: 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 is able to trap after raining.

For example, 
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

The above elevation map 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. Thanks Marcos for contributing this image!

自己一直在想着从左到右，并且用栈的思想去解，但是这样的话分析起来就很麻烦，而且条件很多。 下面是某位高手的解法，非常简洁，而且这种思想根本不需要考虑太多复杂的判断

class Solution {
public:
    int trap(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        int maxHeight = 0;
        vector<int> leftArr(n);
        
        for (int i = 0; i < n; ++i)
        {
            leftArr[i] = maxHeight;
            if (A[i] > maxHeight)
                maxHeight = A[i];
        }
        
        maxHeight = 0;
        vector<int> rightArr(n);
        for (int i = n - 1; i >= 0; --i)
        {
            rightArr[i] = maxHeight;
            if (A[i] > maxHeight)
                maxHeight = A[i];
        }
        
        int water = 0;
        for (int i = 0; i < n; ++i)
        {
            int temp = min(leftArr[i], rightArr[i]) - A[i];
            
            if (temp > 0)
                water += temp;
        }
        
        return water;
    }
};