Saturday, October 15, 2016

LeetCode 419 - Battleships in a Board


https://leetcode.com/problems/battleships-in-a-board/
Given an 2D board, count how many different 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.
  • Battleships come in different sizes.
  • At least one space of horizontal or vertical separates between two battleships - no adjacent battleships will be given.
Example:
X..X
...X
...X
In the above board there are 2 battleships.

X.
https://discuss.leetcode.com/topic/62970/simple-java-solution
Going over all cells, marking only those that are the "first" cell of the battleship. First cell will be defined as the most top-left cell. We can check for first cells by only counting cells that do not have an 'X' to the left and do not have an 'X' above them.

    public int countBattleships(char[][] board) {
        int m = board.length;
        if (m==0) return 0;
        int n = board[0].length;
        
        int count=0;
        
        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++) {
                if (board[i][j] == '.') continue;
                if (i > 0 && board[i-1][j] == 'X') continue;
                if (j > 0 && board[i][j-1] == 'X') continue;
                count++;
            }
        }
        
        return count;
    }
https://discuss.leetcode.com/topic/64027/share-my-7-line-code-1-line-core-code-3ms-super-easy
https://discuss.leetcode.com/topic/65418/o-n-2-time-and-o-1-space-without-modifying-the-board
    public int countBattleships(char[][] board) {
        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' && (i == 0 || board[i-1][j] == '.') && (j == 0 || board[i][j-1] == '.')) {
                    count++;
                }
            }
        }
        return count;
    }

http://bookshadow.com/weblog/2016/10/13/leetcode-battleships-in-a-board/
由于board中的战舰之间确保有'.'隔开,因此遍历board,若某单元格为'X',只需判断其左边和上边的相邻单元格是否也是'X'。
如果左邻居或者上邻居单元格是'X',则说明当前单元格是左边或者上边战舰的一部分;
否则,令计数器+1

X.
解法II FloodFill
遍历board,用DFS(深度优先搜索)对每一个'X'位置进行探索与标记,同时进行计数。
def countBattleships(self, board): """ :type board: List[List[str]] :rtype: int """ vs = set() h = len(board) w = len(board[0]) if h else 0 def dfs(x, y): for dx, dy in zip((1, 0, -1, 0), (0, 1, 0, -1)): nx, ny = x + dx, y + dy if 0 <= nx < h and 0 <= ny < w: if (nx, ny) not in vs and board[nx][ny] == 'X': vs.add((nx, ny)) dfs(nx, ny) ans = 0 for x in range(h): for y in range(w): if (x, y) not in vs and board[x][y] == 'X': ans += 1 vs.add((x, y)) dfs(x, y) return ans



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