计算战舰的个数 Battleships in a Board

2019独角兽企业重金招聘Python工程师标准>>>

问题：

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?

解决：

① 求战舰的个数，所谓的战舰就是只能是一行或者一列，不能有拐弯。本题限定了不会有相邻的两个战舰的存在，有了这一点限制，那么我们只需要遍历一次二维数组就行了，只要找出战舰的起始点。所谓的战舰起始点，就是指一条军舰上最左边的那个‘X’或者最上面的那个‘X’。

class Solution { //5ms
    public int countBattleships(char[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) return 0;
        int count = 0;
        int m = board.length;
        int n = board[0].length;
        for (int i = 0;i < m;i ++){
            for (int j = 0;j < n;j ++){
                if (board[i][j] == '.' || (i > 0 && board[i - 1][j] == 'X') || (j > 0 && board[i][j - 1] == 'X')){
                    continue;
                }
                count ++;
            }
        }
        return count;
    }
}

class Solution {//4ms
    public int countBattleships(char[][] board) {
        if(board.length == 0) return 0;
        int count = 0;
        for(int i = 0; i < board.length; i ++) {
            for(int j = 0; j < board[0].length; j ++) {
                if(board[i][j] == 'X') {
                    if(i > 0 && board[i - 1][j] == 'X') continue;
                    if(j > 0 && board[i][j - 1] == 'X') continue;
                    count ++;
                }
            }
        }
        return count;
    }
}

转载于:https://my.oschina.net/liyurong/blog/1601475