problem: 0079 word search

0079 Word Search/0079 Word Search.md

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+# 0079. Word Search
+
+* Difficulty: medium
+* Link: https://leetcode.com/problems/word-search/
+* Topics: Array-String, Matrix, Backtracking
+* highlight: 暴力解
+
+# Clarification
+
+1. Check the inputs and outputs
+    - INPUT:List[List[int]]
+    - OUTPUT:boolean
+
+# Naive Solution
+
+### Thought Process
+
+1. 每個 node 去比對
+- Implement
+
+    ```python
+    class Solution:
+        def exist(self, board: List[List[str]], word: str) -> bool:
+
+            if not board or not board[0] or not word:
+                return False
+
+            rows = len(board)
+            columns = len(board[0])
+            lenW = len(word)
+            visited = set()
+
+            def isValid(row, col):
+                return ((0 <= row < rows) and (0 <= col < columns))
+
+            def searchWords(r, c, i):
+                if word[i] == board[r][c] and i == lenW-1:
+                    return True
+
+                visited.add((r,c))
+                directions = [(-1,0), (0,1), (1,0), (0,-1)]
+                for dx, dy in directions:
+                    newR, newC = r + dx, c + dy
+                    if not isValid(newR, newC) or not board[newR][newC] or (newR,newC) in visited: continue
+                    if word[i+1] == board[newR][newC] and searchWords(newR, newC, i+1):
+                        return True
+                visited.remove((r, c))
+                return False
+
+            for r in range(rows):
+                for c in range(columns):
+                    if board[r][c] == word[0] and searchWords(r, c, 0):
+                        return True
+            return False
+    ```
+
+
+### Complexity
+
+- Time complexity:O(N*M) * 4^len(word)
+- Space complexity:
+
+### Note
+
+- `r in range(R)` 這個的效能很低，要進行判斷是還是使用 `0 <= r < R`
+    - 因為這個而 time Limit exceeded