529

Author: ProHiryu

DFS/traditionalDFS/529.md

@@ -0,0 +1,62 @@
+## 529 Minesweeper
+
+#### Description
+
+[link](https://leetcode.com/problems/minesweeper/)
+
+---
+
+#### Solution
+
+**这里讲一下DFS的时间复杂度判断：如果是最坏情况的话，首先因为初始DFS的循环是O(M + N)，每个点的最大遍历长度是O(MN)，这就是最坏情况的时间复杂度**
+
+---
+
+#### Code
+
+> 最坏情况下是O(n^2)
+
+```python
+class Solution:
+    def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
+        if not board:
+            return []
+
+        m, n = len(board), len(board[0])
+        i, j = click[0], click[1]
+
+        # If a mine ('M') is revealed, then the game is over - change it to 'X'.
+        if board[i][j] == 'M':
+            board[i][j] = 'X'
+            return board
+
+        # run dfs to reveal the board
+        self.dfs(board, i, j)
+        return board
+
+    def dfs(self, board, i, j):
+        if board[i][j] != 'E':
+            return
+
+        m, n = len(board), len(board[0])
+        directions = [(-1,-1), (0,-1), (1,-1), (1,0), (1,1), (0,1), (-1,1), (-1,0)]
+
+        mine_count = 0
+
+        for d in directions:
+            ni, nj = i + d[0], j + d[1]
+            if 0 <= ni < m and 0 <= nj < n and board[ni][nj] == 'M':
+                mine_count += 1
+
+        if mine_count == 0:
+            board[i][j] = 'B'
+        else:
+            board[i][j] = str(mine_count)
+            return
+
+        # if not mine_count nor mine, is Blank for spread dfs
+        for d in directions:
+            ni, nj = i + d[0], j + d[1]
+            if 0 <= ni < m and 0 <= nj < n:
+                self.dfs(board, ni, nj)
+```