class Solution(object):
def countBattleships(self, board):
"""
:type board: List[List[str]]
:rtype: int
"""
if not board or len(board[0])==0: return 0
count=1 if board[0][0]=='X' else 0
for i in range(1,len(board)):
if board[i][0]=='X' and board[i-1][0]=='.':
count+=1
for j in range(1,len(board[0])):
if board[0][j]=='X' and board[0][j-1]=='.':
count+=1
for i in range(1,len(board)):
for j in range(1,len(board[0])):
if board[i][j]=='X' and board[i][j-1]=='.' \
and board[i-1][j]=='.':
count+=1
return count
419. Battleships in a Board
最后编辑于 :
©著作权归作者所有,转载或内容合作请联系作者
- 文/潘晓璐 我一进店门,熙熙楼的掌柜王于贵愁眉苦脸地迎上来,“玉大人,你说我怎么就摊上这事。” “怎么了?”我有些...
- 文/花漫 我一把揭开白布。 她就那样静静地躺着,像睡着了一般。 火红的嫁衣衬着肌肤如雪。 梳的纹丝不乱的头发上,一...
- 文/苍兰香墨 我猛地睁开眼,长吁一口气:“原来是场噩梦啊……” “哼!你这毒妇竟也来了?” 一声冷哼从身侧响起,我...
推荐阅读更多精彩内容
- 题目描述 Given an 2D board, count how many battleships are in...
- 问题: Given an 2D board, count how many different battleshi...
- 思路(代码优化了该思路) 初始化:count = 0 遍历二维数组:遇到'X'时判断当前块的上方和左方存在'X'与...