Create Battleships in a Board.py

Author: sugia

Battleships in a Board.py

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+'''
+Given an 2D board, count how many battleships are in it. The battleships are represented with 'X's, empty slots are represented with '.'s. You may assume the following rules:
+You receive a valid board, made of only battleships or empty slots.
+Battleships can only be placed horizontally or vertically. In other words, they can only be made of the shape 1xN (1 row, N columns) or Nx1 (N rows, 1 column), where N can be of any size.
+At least one horizontal or vertical cell separates between two battleships - there are no adjacent battleships.
+Example:
+X..X
+...X
+...X
+In the above board there are 2 battleships.
+Invalid Example:
+...X
+XXXX
+...X
+This is an invalid board that you will not receive - as battleships will always have a cell separating between them.
+Follow up:
+Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?
+'''
+
+class Solution(object):
+    def countBattleships(self, board):
+        """
+        :type board: List[List[str]]
+        :rtype: int
+        """
+        res = 0
+        for i in xrange(len(board)):
+            for j in xrange(len(board[i])):
+                if board[i][j] == 'X':
+                    res += 1
+                    self.find(board, i, j)
+                    
+        return res
+    
+    def find(self, board, x, y):
+        vec = [(x, y)]
+        board[x][y] = '.'
+        
+        while vec:
+            x, y = vec.pop()
+            for i, j in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
+                if 0 <= x + i < len(board) and 0 <= y + j < len(board[x+i]):
+                    if board[x+i][y+j] == 'X':
+                        vec.append((x+i, y+j))
+                        board[x+i][y+j] = '.'
+