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  • LeetCode

    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 {
          public int countBattleships(char[][] board) {
              if (board == null)
                  return 0;
              int cnt = 0;
              for (int i=0; i<board.length; i++) {
                  for (int j=0; j<board[i].length; j++) {
                      if (board[i][j] == 'X' && (i==0 || board[i-1][j]=='.') && (j==0 || board[i][j-1]=='.')) {
                          cnt ++;
                      }
                  }
              }
              return cnt;
          }
      }
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  • 原文地址:https://www.cnblogs.com/wxisme/p/7444010.html
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