[LeetCode] 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?

使用DFS来遍历board

如果遇到X并标记，并沿水平和垂直方向继续遍历，

如果遇到.直接返回。

class Solution {
public:
    int countBattleships(vector<vector<char>>& board) {
        if (board.empty())
            return 0;
        // Count battleships
        int res = 0;
        int m = board.size(), n = board[0].size();
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        // Traversal board and judge battleships.
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (!visited[i][j] && board[i][j] == 'X') {
                    dfs(board, visited, i, j);
                    res += 1;
                }
            }
        }
        return res;
    }
    
    void dfs(vector<vector<char>>& board, vector<vector<bool>>& visited, int i, int j) {
        if (visited[i][j] == true)
            return;
        visited[i][j] = true;
        if (board[i][j] == '.') {
            return;
        }
        else {
            if (i + 1 < board.size() && board[i + 1][j] == 'X') 
                dfs(board, visited, i + 1, j);
            if (j + 1 < board[0].size() && board[i][j + 1] == 'X') 
                dfs(board, visited, i, j + 1);
        }
        return;
    }
};
// 9 ms