島嶼數量
“島嶼數量”問題的詳細指南和 Python 實作。
1. 學習
「島嶼數量」問題是圖表部分的關鍵挑戰。
此實作著重於 Python 中的簡單層級邏輯。
在我們提供的解決方案中,我們優先考慮技術準確性和程式碼可讀性。
2. Real-World Applications
3. Visual Intuition
可視化島嶼數量的邏輯流程。
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
仔細閱讀島嶼數量的問題陳述。
2. Formulate brute force
起草一個簡單的迭代解決方案。
3. Identify inefficiency
尋找冗餘計算。
4. Optimize search path
使用散列或排序來加速該過程。
5. Final Implementation
清理生產標準代碼。
問題陳述
給定一個 m x n 二維二進位網格,它代表「1」(陸地)和「0」(水)的地圖,返回島嶼的數量。
島嶼四面環水,相鄰陸地水平或垂直連接而成。您可以假設網格的所有四個邊緣都被水包圍。
寫一個函數 numIslands(grid: List[List[str]]) -> int。
- •m == len(grid)
- •n == len(grid[i])
- •1 <= m, n <= 300
- •grid[i][j] is '0' or '1'
範例
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
- 空輸入結構
- 單元素輸入
- 大數值範圍
準備好解決了嗎?
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面試見解和變化
複雜度分析分解
為什麼時間: Directly evaluates all possibilities.
為什麼選擇太空: Uses standard local memory.
為什麼時間: Optimized paths reduce total operations.
為什麼選擇太空: May trade memory for speed.
最佳化解決方案Python程式碼
最佳化解決方案Python程式碼
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暴力破解代碼(劇透保護)
暴力破解代碼(劇透保護)
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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