每日一题，每日一练。11车的可用捕获量（半夜两点在棋盘上左右横跳），

在一个 8 x 8 的棋盘上，有一个白色车（rook）。也可能有空方块，白色的象（bishop）和黑色的卒（pawn）。它们分别以字符
“R”，“.”，“B” 和 “p” 给出。大写字符表示白棋，小写字符表示黑棋。

车按国际象棋中的规则移动：它选择四个基本方向中的一个（北，东，西和南），然后朝那个方向移动，直到它选择停止、到达棋盘的边缘或移动到同一方格来捕获该方格上颜色相反的卒。另外，车不能与其他友方（白色）象进入同一个方格。

返回车能够在一次移动中捕获到的卒的数量。

示例 1：

输入：[[".",".",".",".",".",".",".","."],[".",".",".",“p”,".",".",".","."],[".",".",".",“R”,".",".",".",“p”],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",“p”,".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释： 在本例中，车能够捕获所有的卒。

示例 2：

输入：[[".",".",".",".",".",".",".","."],[".",“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”,".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：0
解释： 象阻止了车捕获任何卒。

示例 3：

输入：[[".",".",".",".",".",".",".","."],[".",".",".",“p”,".",".",".","."],[".",".",".",“p”,".",".",".","."],[“p”,“p”,".",“R”,".",“p”,“B”,"."],[".",".",".",".",".",".",".","."],[".",".",".",“B”,".",".",".","."],[".",".",".",“p”,".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释： 车可以捕获位置 b5，d6 和 f5 的卒。

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提示：

board.length == board[i].length == 8 board[i][j] 可以是 ‘R’，’.’，‘B’ 或 ‘p’
只有一个格子上存在 board[i][j] == ‘R’

我承认我一开始找车完以后写了个深搜（淦），后来发现这题练得不是深搜，就是遍历写法==于是写了个四个循环（写的更多了，淦）

找到车的位置，向四个方向进行深搜，搜到棋或边界退出，如果搜到卒先+1再退出

代码如下

class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        flag=1
        co=0
        row=0
        li=0
        for i in range(len(board)):
            for ii in range(len(board[0])):
                if(board[i][ii]=="R"):
                    row=i
                    li=ii
                    flag=0
                    break
            if(flag==0):
                break
        for c in range(li+1,len(board)):
            try:
                if(board[row][c]=="p"):
                    co=co+1
                    break
                elif(board[row][c]=="B"):
                    break
            except IndexError:
                    break
        for c in range(li-1,-1,-1):
            try:
                if(board[row][c]=="p"):
                    co=co+1
                    break
                elif(board[row][c]=="B"):
                    break
            except IndexError:
                    break
        for d in range(row+1,len(board)):
            try:
                if(board[d][li]=="p"):
                    co=co+1
                    break
                elif(board[d][li]=="B"):
                    break
            except IndexError:
                    break
        for d in range(row,-1,-1):
            try:
                if(board[d][li]=="p"):
                    co=co+1
                    break
                elif(board[d][li]=="B"):
                    break
            except IndexError:
                    break
        return co

内存：我真的一滴都不剩了