週邊地區
「週邊區域」問題的詳細指南和 Python 實作。
1. 學習
「包圍區域」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
可視化周圍區域的邏輯流程。
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
仔細閱讀週邊地區的問題陳述。
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
給定一個包含「X」和「O」的 m x n 矩陣板，捕捉所有由「X」在 4 方向包圍的區域。
透過將包圍區域中的所有“O”翻轉為“X”來捕捉區域。
寫一個函數 solve(board: List[List[str]]) -> List[List[str]] 傳回修改後的板。
- •m == len(board)
- •n == len(board[i])
- •1 <= m, n <= 200
- •board[i][j] is 'X' or 'O'
範例
board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
Surrounded regions should not be on the border, which means any 'O' on the border of the board is not flipped to 'X'. Any 'O' that is connected to a border 'O' is also not flipped.
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 solve_opt(board):
ROWS, COLS = len(board), len(board[0])
def capture(r, c):
if r < 0 or r == ROWS or c < 0 or c == COLS or board[r][c] != "O": return
board[r][c] = "T"
capture(r + 1, c); capture(r - 1, c); capture(r, c + 1); capture(r, c - 1)
for r in range(ROWS):
for c in range(COLS):
if board[r][c] == "O" and (r in [0, ROWS - 1] or c in [0, COLS - 1]):
capture(r, c)
for r in range(ROWS):
for c in range(COLS):
if board[r][c] == "O": board[r][c] = "X"
for r in range(ROWS):
for c in range(COLS):
if board[r][c] == "T": board[r][c] = "O"暴力破解代碼（劇透保護）
暴力破解代碼（劇透保護）
def solve_brute(board):
ROWS, COLS = len(board), len(board[0])
def dfs(r, c, visited):
if r < 0 or r == ROWS or c < 0 or c == COLS: return False
if board[r][c] == 'X' or (r, c) in visited: return True
visited.add((r, c))
return dfs(r+1, c, visited) and dfs(r-1, c, visited) and dfs(r, c+1, visited) and dfs(r, c-1, visited)
for r in range(ROWS):
for c in range(COLS):
if board[r][c] == 'O':
visited = set()
if dfs(r, c, visited):
for vr, vc in visited: board[vr][vc] = 'X'Algorithm Pattern Checklist
When dealing with Graphs data patterns.
- Are constraints clear?
- Is there a linear or logarithmic optimization possible?
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