LeetCode Battleships in a Board

原题链接在这里：https://leetcode.com/problems/battleships-in-a-board/#/description

题目：

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?

题解：

无论是1*N 还是 N*1型的battleship, 都以左上角的X来标志. 只有遇到X, 并且左边或者上边没有连着的X时才算新的battleship.

Time Complexity: O(mn), m = board.length, n = board[0].length.

Space: O(1).

AC Java:

 1 public class Solution {
 2     public int countBattleships(char[][] board) {
 3         if(board == null || board.length == 0 || board[0].length == 0){
 4             return 0;
 5         }
 6         
 7         int res = 0;
 8         for(int i = 0; i<board.length; i++){
 9             for(int j = 0; j<board[0].length; j++){
10                 if(board[i][j] == '.'){
11                     continue;
12                 }
13                 if(i>0 && board[i-1][j]=='X'){
14                     continue;
15                 }
16                 if(j>0 && board[i][j-1]=='X'){
17                     continue;
18                 }
19                 res++;
20             }
21         }
22         return res;
23     }
24 }

 

转载于:https://www.cnblogs.com/Dylan-Java-NYC/p/6751562.html