leetcode419. Battleships in a Board

题目要求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…XIn the above board there are 2 battleships.Invalid Example:…XXXXX…XThis 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?假设有一个2D板，在板上用X表示战舰，已知板上任意两个战舰体之间一定会用.隔开，因此不会出现两个X相邻的情况。现在要求用O(N)的时间复杂度和O(1)的空间复杂度来完成。思路和代码这题的思路非常清晰，我们只需要判断哪个X是战舰头即可，当我们遇到战舰头时，就将总战舰数加一，其余时候都继续遍历。战舰头即战舰的左侧和上侧没有其它的X。 public int countBattleships(char[][] board) { int count = 0; if(board == null || board.length == 0 || board[0].length == 0) return count; for(int i = 0 ; i<board.length ; i++) { for(int j = 0 ; j<board[i].length ; j++) { if(board[i][j] == ‘X’) { if((i > 0 && board[i-1][j] == ‘X’) || (j > 0 && board[i][j-1] == ‘X’)) { continue; } count++; } } } return count; }