Snap 面试中的几何算法题。本文通过建筑物周长计算问题，展示如何处理二维网格上的几何问题csvosupport* 助你掌握几何算法核心技巧
📋 问题重新定义
给定一个二维网格，其中 1 表示建筑物，0 表示空地。建筑物可能连接形成不规则形状。计算所有建筑物的外围周长（只计算与空地或边界接触的边）
*示例
Input: grid = [
[1,1,0],
[1,1,0],
[0,0,1]
]
Output: 12
🎯 核心思路
- 边界识别 - 找出所有与空地接触的边
- 去重处理 - 避免重复计算共享
- 边界情况 - 处理网格边缘
- 优化策略 - 减少不必要的遍历
💡 解题方法（csvosupport 创新解法
方法一：直接计数法
def calculate_perimeter(grid):
if not grid or not grid[0]:
return 0
rows, cols = len(grid), len(grid[0])
perimeter = 0
for i in range(rows):
for j in range(cols):
if grid[i][j] == 1:
# 检查四个方
#
if i == 0 or grid[i-1][j] == 0:
perimeter += 1
#
if i == rows-1 or grid[i+1][j] == 0:
perimeter += 1
#
if j == 0 or grid[i][j-1] == 0:
perimeter += 1
#
if j == cols-1 or grid[i][j+1] == 0:
perimeter += 1
return perimeter
时间复杂度： O(m × n) 空间复杂度： O(1)
方法二：数学优化
csvosupport 提供的优化思路
def calculate_perimeter_optimized(grid):
rows, cols = len(grid), len(grid[0])
buildings = 0
shared_edges = 0
for i in range(rows):
for j in range(cols):
if grid[i][j] == 1:
buildings += 1
# 检查右边和下边的共享边
if j < cols - 1 and grid[i][j+1] == 1:
shared_edges += 1
if i < rows - 1 and grid[i+1][j] == 1:
shared_edges += 1
# 每个建筑贡献4条边，减去共享边
return buildings * 4 - shared_edges * 2
🚀 扩展问题
1. 计算内部空洞的周
def calculate_with_holes(grid):
def dfs(i, j, visited):
if (i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or
(i, j) in visited or grid[i][j] == 1):
return 0
visited.add((i, j))
perimeter = 0
# 检查四个方
for di, dj in [(0,1), (0,-1), (1,0), (-1,0)]:
ni, nj = i + di, j + dj
if (0 <= ni < len(grid) and 0 <= nj < len(grid[0]) and
grid[ni][nj] == 1):
perimeter += 1
else:
perimeter += dfs(ni, nj, visited)
return perimeter
# 从边界的空地开始DFS，找到外围周
visited = set()
outer_perimeter = 0
# 遍历边界
for i in range(len(grid)):
if grid[i][0] == 0:
outer_perimeter += dfs(i, 0, visited)
if grid[i][len(grid[0])-1] == 0:
outer_perimeter += dfs(i, len(grid[0])-1, visited)
for j in range(len(grid[0])):
if grid[0][j] == 0:
outer_perimeter += dfs(0, j, visited)
if grid[len(grid)-1][j] == 0:
outer_perimeter += dfs(len(grid)-1, j, visited)
return outer_perimeter
2. 3D 建筑物表面积
def calculate_surface_area_3d(grid):
"""
grid[i][j] 表示该位置建筑物的高
"""
if not grid:
return 0
rows, cols = len(grid), len(grid[0])
surface_area = 0
for i in range(rows):
for j in range(cols):
if grid[i][j] > 0:
# 顶部和底
surface_area += 2
# 四个侧面
for di, dj in [(0,1), (0,-1), (1,0), (-1,0)]:
ni, nj = i + di, j + dj
neighbor_height = 0
if 0 <= ni < rows and 0 <= nj < cols:
neighbor_height = grid[ni][nj]
# 当前建筑高于邻居的部分贡献表面积
if grid[i][j] > neighbor_height:
surface_area += grid[i][j] - neighbor_height
return surface_area
💼 csvosupport 助力
几何思维 - 将问题转化为几何计算 优化策略 - 数学方法减少计算 扩展能力 - 3D 和复杂场 代码实现 - 清晰的逻辑结构
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*标签 #Snap #几何算法 #网格遍历 #VO辅助 #面试辅助 #一亩三分地
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