岛屿数量
“岛屿数量”问题的详细指南和 Python 实现。
1. 学习
“岛屿数量”问题是图表部分的一个关键挑战。
此实现侧重于 Python 中的简单级逻辑。
在我们提供的解决方案中,我们优先考虑技术准确性和代码可读性。
2. Real-World Applications
3. Visual Intuition
可视化岛屿数量的逻辑流程。
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
仔细阅读岛屿数量的问题陈述。
2. Formulate brute force
起草一个简单的迭代解决方案。
3. Identify inefficiency
寻找冗余计算。
4. Optimize search path
使用散列或排序来加速该过程。
5. Final Implementation
清理生产标准代码。
问题陈述
给定一个 m x n 二维二进制网格,它代表“1”(陆地)和“0”(水)的地图,返回岛屿的数量。
岛屿四面环水,相邻陆地水平或垂直连接而成。您可以假设网格的所有四个边缘都被水包围。
编写一个函数 numIslands(grid: List[List[str]]) -> int。
- •m == len(grid)
- •n == len(grid[i])
- •1 <= m, n <= 300
- •grid[i][j] is '0' or '1'
示例
grid = [["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"]]
1
There is a single island consisting of all connected '1's starting from top-left.
grid = [["1","1","0","0","0"],["1","1","0","0","0"],["0","0","1","0","0"],["0","0","0","1","1"]]
3
There are three distinct islands separated by '0's.
Need a Hint?
Edge Cases to Watch
- 空输入结构
- 单元素输入
- 大数值范围
准备好解决了吗?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
面试见解和变化
复杂性分析分解
为什么时间: Directly evaluates all possibilities.
为什么选择太空: Uses standard local memory.
为什么时间: Optimized paths reduce total operations.
为什么选择太空: May trade memory for speed.
优化解决方案Python代码
优化解决方案Python代码
def num_islands_opt(grid: list[list[str]]) -> int:
if not grid: return 0
rows, cols = len(grid), len(grid[0])
islands = 0
def dfs(r, c):
if r < 0 or c < 0 or r >= rows or c >= cols or grid[r][c] == "0":
return
grid[r][c] = "0"
dfs(r + 1, c)
dfs(r - 1, c)
dfs(r, c + 1)
dfs(r, c - 1)
for r in range(rows):
for c in range(cols):
if grid[r][c] == "1":
dfs(r, c)
islands += 1
return islands暴力破解代码(剧透保护)
暴力破解代码(剧透保护)
def num_islands_brute(grid: list[list[str]]) -> int:
if not grid: return 0
rows, cols = len(grid), len(grid[0])
visited = set()
islands = 0
def bfs(r, c):
q = [(r, c)]
visited.add((r, c))
while q:
row, col = q.pop(0)
directions = [[1,0], [-1,0], [0,1], [0,-1]]
for dr, dc in directions:
nr, nc = row + dr, col + dc
if 0 <= nr < rows and 0 <= nc < cols and grid[nr][nc] == "1" and (nr, nc) not in visited:
q.append((nr, nc))
visited.add((nr, nc))
for r in range(rows):
for c in range(cols):
if grid[r][c] == "1" and (r, c) not in visited:
bfs(r, c)
islands += 1
return islandsAlgorithm Pattern Checklist
When dealing with Graphs data patterns.
- Are constraints clear?
- Is there a linear or logarithmic optimization possible?
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