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  • 419. Battleships in a Board

    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:
        def countBattleships(self, board):
            """
            :type board: List[List[str]]
            :rtype: int
            """
            res = 0
            m,n = len(board),len(board[0])
            for i in range(m):
                for j in range(n):
                    if board[i][j]=='X':
                        if not ((i-1>=0 and board[i-1][j]=='X') or (j-1>=0 and board[i][j-1]=='X')):
                            res += 1
            return res
    
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  • 原文地址:https://www.cnblogs.com/bernieloveslife/p/9830280.html
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