Trapping Rain Water

DP.

O(n) solution.

对于每个bar，找出其左边与右边的最高高度， 可以hold的water = min(max_left, max_right) - height

    int trap(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int maxHeight = 0;
        vector<int> left(n);
        for(int i=0;i<n;i++)
        {
            left[i] = maxHeight;
            maxHeight = max(maxHeight,A[i]);
        }
        
        maxHeight = 0;
        vector<int> right(n);
        for(int i=n-1;i>=0;i--)
        {
            right[i] = maxHeight;
            maxHeight = max(maxHeight,A[i]);
        }
        int water = 0;
        for(int i=0;i<n;i++)
        {
            int height = min(left[i],right[i]) - A[i];
            if(height<0)
                height = 0;
            water += height;
        }
        return water;
        
    }