On a N * Ngrid, we place some 1 * 1 * 1cubes that are axis-aligned with the x, y, and z axes.

Each value v = grid[i][j] represents a tower of v cubes placed on top of grid cell(i, j).

Now we view the projection of these cubes onto the xy, yz, and zx planes.

A projection is like a shadow, that maps our 3 dimensional figure to a 2 dimensional plane.

Here, we are viewing the "shadow" when looking at the cubes from the top, the front, and the side.

Return the total area of all three projections.

Example:

Input: [[1,2],[3,4]]
Output: 17
Explanation: 
Here are the three projections ("shadows") of the shape made with each axis-aligned plane.

这道题本来就老老实实地计算3个方向的投影就可以了，直接计算需要储存每一行和每一列的投影，这样的话需要额外的空间。但是有一个可以优化的原因是题目中给定了这个矩阵是N * N的，所以行列的遍历可以同时进行。如果是更一般化的M * N的矩阵，则至少需要额外的O(M)或O(N)的空间

class Solution {
    public int projectionArea(int[][] grid) {
        if(grid == null || grid.length == 0 || grid[0] == null || grid[0].length == 0)
            return 0;

        int res = 0;
        int bottom = 0;
        int n = grid.length;

        for(int i = 0; i < n; i++) {
            int row = 0;
            int col = 0;
            for(int j = 0; j < n; j++) {
                if(grid[i][j] != 0)
                    res++;
                row = Math.max(row, grid[i][j]);
                col = Math.max(col, grid[j][i]);
            }

            res += row;
            res += col;
        }

        return res;
    }
}