Short Python Solution


  • 0
    W

    For each X in the board, we just check if the square above it is outside of the grid or is a dot. This works because any X with another X above or to the left of it is a continuation of a previously counted battleship.

    class Solution(object):
        def countBattleships(self, board):
            total = 0
            for i in xrange(len(board)):
                for j in xrange(len(board[0])):
                    total += (board[i][j] == "X") and 
                             (i == 0 or board[i-1][j] == ".") and 
                             (j == 0 or board[i][j-1] == ".")
            return total
    

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