LeetCode 892. Surface Area of 3D Shapes (Java版; Easy)

朱雀 2023-07-18 09:11 43阅读 0赞

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## LeetCode 892. Surface Area of 3D Shapes (Java版; Easy) ##

#### 题目描述 ####

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

#### 第一次做; 核心: 1) 一个柱子一个柱子的考虑 2) 考虑贴合面积时, 不要考虑四周, 只需要考虑\[i-1\]\[j\]和\[i\]\[j-1\], 因为遍历的时候是按顺序遍历的, 所以只考虑两个相邻的坐标不出出现少考虑的情况,也不会出现重复考虑的情况 ####

class Solution { 
        public int surfaceArea(int[][] grid) { 
            int n = grid.length;
            int res = 0;
            for(int i=0; i<n; i++){ 
                for(int j=0; j<n; j++){ 
                    if(grid[i][j] > 0){ 
                        res += 4*grid[i][j] + 2;
                    }
                    //核心: 如何考虑相邻? 不用考虑四周, 只需要考虑(i-1,j)和(i,j-1)
                    //不需要考虑grid[i-1][j]是否大于0, 不会影响结果; 考虑了也行
                    if(i-1>=0 && grid[i-1][j]>0){ 
                        res -= 2 * Math.min(grid[i][j], grid[i-1][j]);
                    }
                    if(j-1>=0 && grid[i][j-1]>0){ 
                        res -= 2 * Math.min(grid[i][j], grid[i][j-1]);
                    }
                }
            }
            return res;
        }
    }

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