419.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) orNx1
(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?
Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?
思路:题目意思大致是一个二维数组由’.’与’X’组成,’X’表示战舰,战舰规模可以是多个’X’组成,例如:
...X X.X. X..X X... .X.X (3个) .X.. (3个)
可以是1*N或者N*1组成,计算战舰数量。可通过双重循环,找到 ‘X’ 的位置,sum++,如果前或上有 ‘X’ ,sum–。
代码如下:
public class BattleshipsInABoard {
public int countBattleships(char[][] board) {
int sum = 0;
for (int i=0;i<board.length;i++){
for (int j=0;j<board[i].length;j++){
if (board[i][j]=='X'){
sum++;
if (i>0 && board[i-1][j]=='X')sum--;
if (j>0 && board[i][j-1]=='X')sum--;
}
}
}
return sum;
}
}
Runtime: 4 ms