[leetcode]-892. Surface Area of 3D Shapes（C语言）

On a N * N grid, we place some 1 * 1 * 1 cubes.

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

Return the total surface area of the resulting shapes.

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

Note:

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

int surfaceArea(int** grid, int gridRowSize, int *gridColSizes) {
    int i,j,k,l,sum=0;
    for(i=0;i<gridRowSize;i++)
    {
        for(j=0;j<*gridColSizes;j++)
        {
            if(grid[i][j]==0)
                continue;
            else
            {
                sum+=2+4*grid[i][j];
                if(j-1>=0)
                {
                    if(grid[i][j-1]>=grid[i][j])
                        sum-=grid[i][j];
                    else
                        sum-=grid[i][j-1];
                }
                if(j+1<*gridColSizes)
                {
                    if(grid[i][j+1]>=grid[i][j])
                        sum-=grid[i][j];
                    else
                        sum-=grid[i][j+1];
                }
                if(i-1>=0)
                {
                    if(grid[i-1][j]>=grid[i][j])
                        sum-=grid[i][j];
                    else
                        sum-=grid[i-1][j];
                }
                if(i+1<gridRowSize)
                {
                    if(grid[i+1][j]>=grid[i][j])
                        sum-=grid[i][j];
                    else
                        sum-=grid[i+1][j];
                }
            }
        }
    }
    return sum;
}