Leetcode.892. Surface area of three-dimensional body.

topic

N * N on the grid, we put some cube 1 * 1 * 1.

Each value v = grid [i] [j] denotes the v cube superposed on the corresponding cell (i, j).

Please return to the final body surface area.

Example 1:

Input: [[2]]
Output: 10
Example 2:

Input: [[1,2], [3,4]]
Output: 34
Example 3:

Input: [[1,0], [0,2]]
Output: 16
Example 4:

Input: [[1,1,1], [1,0,1], [1,1,1]]
Output: 32
Example 5:

Input: [[2,2,2], [2,1,2], [2,2,2]]
Output: 46

prompt:

1 <= N <= 50
0 <= grid[i][j] <= 50

Thinking

Initial idea is relatively simple to add this title are marked as easy, assume that the main body of the three-dimensional view, a left side view and a plan view of the sum is multiplied by 2 to result in the article "hollow" state (4,5 Example can not accurately calculate the surface area under). So think of other methods: Imagine that you walk on this three-dimensional body, from left to right, top to bottom traveled that is the surface area of ​​the body. To achieve the following:

class Solution {
public:
    int surfaceArea(vector<vector<int>>& grid) {
        int N=grid.size();
        if(N==0)
            return 0;
        int ispan=0, jspan=0;
        for(int i=0; i<N; i++){
            for(int j=0; j<N; j++){
                if(j==0)
                    jspan+=grid[i][j];
                else
                    jspan+=abs(grid[i][j]-grid[i][j-1]);
                if(grid[i][j] != 0)
                    jspan+=1;
                if(j == N-1)
                    jspan+=grid[i][j];

                if(i == 0)
                    ispan+=grid[i][j];
                else
                    ispan+=abs(grid[i][j]-grid[i-1][j]);
                if(grid[i][j] != 0)
                    ispan+=1;//加了两次，正好对应上面和下面
                if(i == N-1)
                    ispan+=grid[i][j];
            }
        }

        return ispan+jspan;
    }
};