leetcode-419-Battleships-in-a-Board

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描述

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:

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X..X
...X
...X

In the above board there are 2 battleships.

Invalid Example:

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3
...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?

分析

这道题跟 695. Max Area of Island 原理基本一样，只不过那道题是求最大的连通图，而这道题是求连通图的个数，代码改几行就能用。用 DFS 遍历即可。

解决方案1(Java)

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class Solution {
    int[][] steps = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
    public int countBattleships(char[][] board) {
        int result = 0;
        int row = board.length;
        int column = board[0].length;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < column; j++) {
                if (board[i][j] == 'X') {
                    result++;
                    dfs(board, i, j, row, column);
                }
            }
        }
        return result;
    }
    
    private void dfs(char[][] board, int x, int y, int row, int column) {
        if (x < 0 || y < 0 || x >= row || y >= column || board[x][y] == '.') {
            return;
        }
        board[x][y] = '.';
        for (int[] step: steps) {
            dfs(board, x+step[0], y+step[1], row, column);
        }
    }
}

题目来源