题目描述:

题解：

（1）暴力递归

class Solution(object):
    def find(self,i,j,board):
        if(board[i][j]=='.'):
            return      
        board[i][j]='.'
        if( board[i-1][j]=='X' and i-1>0):
            self.find(i-1,j,board)
        if( j+1<len(board[0]) and board[i][j+1]=='X' ):
            self.find(i,j+1,board)
        if( i+1<len(board) and board[i+1][j]=='X'):
            self.find(i+1,j,board)
        if( board[i][j-1]=='X' and j-1>0):
            self.find(i,j-1,board)
    def countBattleships(self, board):
        count = 0
        for i in range(len(board)):
            for j in range(len(board[0])):
                if(board[i][j]=='X'):
                    count+=1
                    self.find(i,j,board)
        return count

（2）遍历扫描

class Solution:
    def countBattleships(self, board: List[List[str]]) -> int:
        ans = 0
        m, n = len(board), len(board[0])
        for i, row in enumerate(board):
            for j, ch in enumerate(row):
                if ch == 'X':
                    row[j] = '.'
                    for k in range(j + 1, n):
                        if row[k] != 'X':
                            break
                        row[k] = '.'
                    for k in range(i + 1, m):
                        if board[k][j] != 'X':
                            break
                        board[k][j] = '.'
                    ans += 1
        return ans

（3）枚举起点

class Solution {
    public int countBattleships(char[][] board) {
        int row = board.length;
        int col = board[0].length;
        int ans = 0;
        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < col; ++j) {
                if (board[i][j] == 'X') {
                    if (i > 0 && board[i - 1][j] == 'X') {
                        continue;
                    }
                    if (j > 0 && board[i][j - 1] == 'X') {
                        continue;
                    }
                    ans++;
                }
            }
        }
        return ans;
    }
}