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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 <= 500 <= grid[i][j] <= 50
在 N * N 的网格上，我们放置一些 1 * 1 * 1 的立方体。
每个值 v = grid[i][j] 表示 v 个正方体叠放在单元格 (i, j) 上。
返回结果形体的总表面积。
示例 1：
输入：[[2]] 输出：10
示例 2：
输入：[[1,2],[3,4]] 输出：34
示例 3：
输入：[[1,0],[0,2]] 输出：16
示例 4：
输入：[[1,1,1],[1,0,1],[1,1,1]] 输出：32
示例 5：
输入：[[2,2,2],[2,1,2],[2,2,2]] 输出：46
提示：
1 <= N <= 500 <= grid[i][j] <= 50
Runtime: 52 ms
Memory Usage: 19.1 MB
1 class Solution { 2 func surfaceArea(_ grid: [[Int]]) -> Int { 3 var res:Int = 0 4 var n:Int = grid.count 5 for i in 0..<n 6 { 7 for j in 0..<n 8 { 9 if grid[i][j] > 0 10 { 11 res += grid[i][j] * 4 + 2 12 } 13 if i > 0 14 { 15 res -= min(grid[i][j], grid[i - 1][j]) * 2 16 } 17 if j > 0 18 { 19 res -= min(grid[i][j], grid[i][j - 1]) * 2 20 } 21 } 22 } 23 return res 24 } 25 }
60ms
1 class Solution { 2 func surfaceArea(_ grid: [[Int]]) -> Int { 3 let n = grid.count 4 var result = 0 5 6 for i in 0..<n { 7 for j in 0..<n { 8 let g = grid[i][j] 9 if g == 0 { 10 continue 11 } 12 result += 2 13 14 if i > 0 { 15 result += max(0, g-grid[i-1][j]) 16 } else { 17 result += g 18 } 19 20 if j > 0 { 21 result += max(0, g-grid[i][j-1]) 22 } else { 23 result += g 24 } 25 26 if i+1 < n { 27 result += max(0, g-grid[i+1][j]) 28 } else { 29 result += g 30 } 31 32 if j+1 < n { 33 result += max(0, g-grid[i][j+1]) 34 } else { 35 result += g 36 } 37 } 38 } 39 return result 40 } 41 }
88ms
1 class Solution { 2 func surfaceArea(_ grid: [[Int]]) -> Int { 3 4 var sum=0,a=0,b=0,c=0,d=0 5 for (i,item) in grid.enumerated() { 6 for (j,subitem) in item.enumerated() { 7 if subitem > 0 { 8 sum += 2 9 a = j < grid[i].count-1 ? grid[i][j+1] : 0 10 b = j > 0 ? grid[i][j-1] : 0 11 c = i > 0 ? grid[i-1][j] : 0 12 d = i < grid.count-1 ? grid[i+1][j] : 0 13 sum += (max(subitem-a, 0)+max(subitem-b, 0)+max(subitem-c, 0)+max(subitem-d, 0)) 14 } 15 } 16 } 17 return sum 18 } 19 }
100ms
1 class Solution { 2 func surfaceArea(_ grid: [[Int]]) -> Int { 3 func getSingleFaceArea(_ arr: [Int]) -> Int { 4 var result = 0 5 //上一个 x 6 var lastX = 0 7 arr.forEach { (x) in 8 //单独一堆的面积 9 let temp = (x == 0 ? 0 : (6 * x) - ((x - 1) * 2)) 10 result += temp 11 let minX = lastX <= x ? lastX : x 12 //减去两堆拼起来以后，遮住的面积 13 result -= 2 * minX 14 lastX = x 15 } 16 return result 17 } 18 19 var result = 0 20 var lastArr: [Int] = [] 21 for i in (0..<grid.count) { 22 let current = grid[i] 23 result += getSingleFaceArea(current) 24 if i != 0 { 25 for i in (0..<current.count) { 26 //减去两排拼起来以后，遮住的面积 27 let x = lastArr[i] <= current[i] ? lastArr[i] : current[i] 28 result -= x * 2 29 } 30 } 31 lastArr = current 32 } 33 return result 34 } 35 }