在乙個 8 x 8 的棋盤上，有乙個白色車（rook）。也可能有空方塊，白色的象（bishop）和黑色的卒（pawn）。它們分別以字元 「r」，「.」，「b」 和 「p」 給出。大寫字元表示白棋，小寫字元表示黑棋。
車按西洋棋中的規則移動：它選擇四個基本方向中的乙個（北，東，西和南），然後朝那個方向移動，直到它選擇停止、到達棋盤的邊緣或移動到同一方格來捕獲該方格上顏色相反的卒。另外，車不能與其他友方（白色）象進入同乙個方格。
返回車能夠在一次移動中捕獲到的卒的數量。
輸入：[[".",".",".",".",".",".",".","."],
[".",".",".",「p」,".",".",".","."],
[".",".",".",「r」,".",".",".",「p」],
[".",".",".",".",".",".",".","."],
[".",".",".",".",".",".",".","."],
[".",".",".",「p」,".",".",".","."],
[".",".",".",".",".",".",".","."],
[".",".",".",".",".",".",".","."]]
輸出：3
解釋：在本例中，車能夠捕獲所有的卒。
輸入：[[".",".",".",".",".",".",".","."],
[".",「p」,「p」,「p」,「p」,「p」,".","."],
[".",「p」,「p」,「b」,「p」,「p」,".","."],
[".",「p」,「b」,「r」,「b」,「p」,".","."],
[".",「p」,「p」,「b」,「p」,「p」,".","."],
[".",「p」,「p」,「p」,「p」,「p」,".","."],
[".",".",".",".",".",".",".","."],
[".",".",".",".",".",".",".","."]]
輸出：0
解釋：象阻止了車捕獲任何卒。
輸入：[[".",".",".",".",".",".",".","."],
[".",".",".",「p」,".",".",".","."],
[".",".",".",「p」,".",".",".","."],
[「p」,「p」,".",「r」,".",「p」,「b」,"."],
[".",".",".",".",".",".",".","."],
[".",".",".",「b」,".",".",".","."],
[".",".",".",「p」,".",".",".","."],
[".",".",".",".",".",".",".","."]]
輸出：3
解釋：車可以捕獲位置 b5，d6 和 f5 的卒。
題目描述的真的挺繞的，簡單說就是車可以向上下左右四個方向走，遇到象就不能再繼續走了，同一方向的車只能捕獲乙個。
簡單方法：首先找到車的位置，然後上下左右四個方向分別搜尋是否有車。對每個方向，若遇到象，則直接退回，一旦捕獲到了乙個車，也退回。
class
solution
(object)
: def numrookcaptures
(self, board)
:"""
:type board: list[list[str]
]:rtype:
int"""
# 找到車r的位置
break_flag = false # 記錄內層迴圈是否break
for i in range(0
,8):
for j in range(0
,8):
if board[i]
[j]==
'r':
m = i
n = j
break_flag = true
break
# 由於只有乙個r，若內層j迴圈break，則外層i也break
if break_flag:
break
count =
0
# 向右搜尋
i = n+
1while i<8:
if board[m]
[i]==
'b':
break
elif board[m]
[i]==
'p':
count +=
1break
i +=
1
# 向左搜尋
i = n-
1while i>0:
if board[m]
[i]==
'b':
break
elif board[m]
[i]==
'p':
count +=
1break
i -=
1
# 向上搜尋
i = m-
1while i>0:
if board[i]
[n]==
'b':
break
elif board[i]
[n]==
'p':
count +=
1break
i -=
1
# 向下搜尋
i = m+
1while i<8:
if board[i]
[n]==
'b':
break
elif board[i]
[n]==
'p':
count +=
1break
i +=
1return count
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