单词搜索
“单词搜索”问题的详细指南和 Python 实现。
1. 学习
“单词搜索”问题是回溯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
可视化单词搜索的逻辑流程。
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
仔细阅读单词搜索的问题陈述。
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 字符板网格和一个字符串单词，如果单词存在于网格中，则返回 true。
该单词可以由顺序相邻的单元格的字母构成，其中相邻的单元格水平或垂直相邻。同一字母单元不得使用多次。
实现函数 exist(board: list, word: str) -> bool。
- •m == board.length
- •n == board[i].length
- •1 <= m, n <= 6
- •1 <= word.length <= 15
- •board and word consists of only lowercase and uppercase English letters
示例
[["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], "ABCCED"
True
The word ABCCED can be traced: A(0,0)->B(0,1)->C(0,2)->C(1,2)->E(2,2)->D(2,1).
[["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], "SEE"
True
The word SEE can be traced: S(1,3)->E(2,3)->E(2,2).
[["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], "ABCB"
False
Cannot trace ABCB without reusing cells.
Need a Hint?
Edge Cases to Watch
- 空输入结构
- 单元素输入
- 大数值范围
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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 exist_opt(board: list[list[str]], word: str) -> bool:
rows, cols = len(board), len(board[0])
path = set()
def dfs(r, c, i):
if i == len(word): return True
if r < 0 or c < 0 or r >= rows or c >= cols or board[r][c] != word[i] or (r, c) in path:
return False
path.add((r, c))
res = dfs(r + 1, c, i + 1) or dfs(r - 1, c, i + 1) or dfs(r, c + 1, i + 1) or dfs(r, c - 1, i + 1)
path.remove((r, c))
return res
from collections import Counter
count = Counter(word)
board_count = Counter(char for row in board for char in row)
for char in count:
if count[char] > board_count[char]: return False
if board_count[word[0]] > board_count[word[-1]]: word = word[::-1]
for r in range(rows):
for c in range(cols):
if dfs(r, c, 0): return True
return False暴力破解代码（剧透保护）
暴力破解代码（剧透保护）
def exist_brute(board: list[list[str]], word: str) -> bool:
rows, cols = len(board), len(board[0])
def dfs(r, c, i, visited):
if i == len(word): return True
if r < 0 or c < 0 or r >= rows or c >= cols or (r, c) in visited or board[r][c] != word[i]:
return False
visited.add((r, c))
res = dfs(r + 1, c, i + 1, visited) or dfs(r - 1, c, i + 1, visited) or dfs(r, c + 1, i + 1, visited) or dfs(r, c - 1, i + 1, visited)
visited.remove((r, c))
return res
for r in range(rows):
for c in range(cols):
if dfs(r, c, 0, set()): return True
return FalseAlgorithm Pattern Checklist
When dealing with Backtracking data patterns.
- Are constraints clear?
- Is there a linear or logarithmic optimization possible?
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