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class Solution {
    private int[][] directions = new int[][] { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 } };

    public int countBattleships(char[][] board) {
        int m = board.length, n = board[0].length;
        boolean[][] visited = new boolean[m][n];
        int ans = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == 'X' && !visited[i][j]) {
                    dfs(board, visited, i, j);
                    ans += 1;
                }
            }
        }
        return ans;
    }

    private void dfs(char[][] board, boolean[][] visited, int row, int col) {
        visited[row][col] = true;
        for (int[] direction : directions) {
            int newRow = row + direction[0];
            int newCol = col + direction[1];
            if (!isOutOfBound(board, newRow, newCol) && board[newRow][newCol] == 'X' && !visited[newRow][newCol]) {
                dfs(board, visited, newRow, newCol);
            }
        }
    }

    private boolean isOutOfBound(char[][] board, int row, int col) {
        return row < 0 || row >= board.length || col < 0 || col >= board[0].length;
    }
}

DFS就是.

时间复杂度: O(m * n) 遍历了整个board.
空间复杂度: O(m * n) 用了visited这个数组.

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

这个做法十分聪明, 只是查每个battleship第一个cell. 我们定义第一个cell为top-left的那个.

时间复杂度和空间复杂度不变.