單字搜尋
“單字搜尋”問題的詳細指南和 Python 實作。
1. 學習
「單字搜尋」問題是回溯部分的關鍵挑戰。
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
可視化單字搜尋的邏輯流程。
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
仔細閱讀單字搜尋的問題陳述。
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 字元板網格和一個字串單詞，如果單字存在於網格中，則傳回 true。
該單字可以由順序相鄰的單元格的字母構成，其中相鄰的單元格水平或垂直相鄰。同一字母單元不得使用多次。
實作函數 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
範例
[["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
- 空輸入結構
- 單元素輸入
- 大數值範圍
準備好解決了嗎？
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
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 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暴力破解代碼（劇透保護）
暴力破解代碼（劇透保護）
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?
PyRun is built and maintained by an independent solo developer. If this helped your interview prep, consider buying a coffee!
推薦的 Python 資源
透過相關的互動式教學、備忘單和程式碼比較來擴展您的知識。
Python 迴圈：For 與 While 迴圈解釋
了解如何使用 Python 循環來迭代資料。透過互動式範例掌握 for 迴圈、while 迴圈、break、continue 和迴圈最佳實務。
如何在 Python 中對列表進行排序（升序和降序）
了解如何在 Python 中使用 sort() 方法和sorted() 函數對清單進行排序。發現自訂鍵排序和逆序範例。
Python 字串方法備忘單
Python 字串操作的完整參考指南。掌握格式化、搜尋、拆分、取代和檢查字串屬性。
Python 與 JavaScript：哪種程式語言最好？
Python 和 JavaScript 的全面比較。探索語法差異、效能、用例（後端與前端）和編碼範例。