999. Available Captures for Rook

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class Solution:     def calcu(self,ele:List[str],loc):         index_p = [-1]*2 #只记录离R最近位置的p和B（左右）         index_B = [-1]*2         left,right = 0,0         for i in range(len(ele)):             if ele[i] is 'p':                 if i < loc:                     index_p[0]=i #左边的取最大值                 else:                     if index_p[1]<0:index_p[1] = i #右边的取最小值             elif ele[i] is 'B':                 if i<loc:                     index_B[0]=i                 else:                     if index_B[1]<0:index_B[1]=i                               p0,p1,b0,b1 = index_p[0],index_p[1],index_B[0],index_B[1] #使用简便的符号标记，方便书写         if p0>=0:#p在R的左边             if b0>=0: #B在R的左边，有值                 if p0<loc and b0<p0:                     left = 1             else: #B左边没有值                 if p0<loc:                     left = 1         if p1>=0:#p在R的右边             if b1>=0:#B在R的右边，有值                 if p1>loc and b1>p1:                     right = 1             else:#B右边没有值                 if p1>loc:                     right = 1                           return left,right                       def numRookCaptures(self, board: List[List[str]]) -> int:         index_R_row = 0         index_R_col = 0         count = 0         col = []         row = []         for a in board:             index_R_row += 1             if 'R' in a:                 for i in range(len(a)):                     if a[i]=='R':                         index_R_col = i                         col = [x[i] for x in board]                         row = a                         break                 break                                     r1,r2 = self.calcu(row,index_R_col)         r3,r4 = self.calcu(col,index_R_row-1)         count = r1+r2+r3+r4         return count

　　40ms,13M

my god，写这个代码花了我半天时间，估计以后都不想看第二遍了。。。。。