Question

419. Battleships in a Board

中文文档

Description

Given an m x n matrix board where each cell is a battleship 'X' or empty '.', return _the number of the battleships on_ board.

Battleships can only be placed horizontally or vertically on board. In other words, they can only be made of the shape 1 x k (1 row, k columns) or k x 1 (k rows, 1 column), where k can be of any size. At least one horizontal or vertical cell separates between two battleships (i.e., there are no adjacent battleships).

 

Example 1:

Input: board = [["X",".",".","X"],[".",".",".","X"],[".",".",".","X"]]
Output: 2

Example 2:

Input: board = [["."]]
Output: 0

 

Constraints:

• m == board.length
• n == board[i].length
• 1 <= m, n <= 200
• board[i][j] is either '.' or 'X'.

 

Follow up: Could you do it in one-pass, using only O(1) extra memory and without modifying the values board?

Solutions

Solution 1

class Solution:
  def countBattleships(self, board: List[List[str]]) -> int:
    m, n = len(board), len(board[0])
    ans = 0
    for i in range(m):
      for j in range(n):
        if board[i][j] == '.':
          continue
        if i > 0 and board[i - 1][j] == 'X':
          continue
        if j > 0 and board[i][j - 1] == 'X':
          continue
        ans += 1
    return ans

class Solution {
  public int countBattleships(char[][] board) {
    int m = board.length, n = board[0].length;
    int ans = 0;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (board[i][j] == '.') {
          continue;
        }
        if (i > 0 && board[i - 1][j] == 'X') {
          continue;
        }
        if (j > 0 && board[i][j - 1] == 'X') {
          continue;
        }
        ++ans;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int countBattleships(vector<vector<char>>& board) {
    int m = board.size(), n = board[0].size();
    int ans = 0;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (board[i][j] == '.') continue;
        if (i > 0 && board[i - 1][j] == 'X') continue;
        if (j > 0 && board[i][j - 1] == 'X') continue;
        ++ans;
      }
    }
    return ans;
  }
};

func countBattleships(board [][]byte) int {
  m, n := len(board), len(board[0])
  ans := 0
  for i := 0; i < m; i++ {
    for j := 0; j < n; j++ {
      if board[i][j] == '.' {
        continue
      }
      if i > 0 && board[i-1][j] == 'X' {
        continue
      }
      if j > 0 && board[i][j-1] == 'X' {
        continue
      }
      ans++
    }
  }
  return ans
}