Battleships in a Board–LeetCode#419

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) 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?
 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

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