周边地区
“周边区域”问题的详细指南和 Python 实现。
1. 学习
“包围区域”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
可视化周围区域的逻辑流程。
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
仔细阅读周边地区的问题陈述。
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
给定一个包含“X”和“O”的 m x n 矩阵板，捕获由“X”在 4 方向包围的所有区域。
通过将包围区域中的所有“O”翻转为“X”来捕获区域。
编写一个函数 solve(board: List[List[str]]) -> List[List[str]] 返回修改后的板。
- •m == len(board)
- •n == len(board[i])
- •1 <= m, n <= 200
- •board[i][j] is 'X' or 'O'
示例
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 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"暴力破解代码（剧透保护）
暴力破解代码（剧透保护）
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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