题目描述：

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

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

示例 3：

输入：[
[".",".",".",".",".",".",".","."],
[".",".",".","p",".",".",".","."],
[".",".",".","p",".",".",".","."],
["p","p",".","R",".","p","B","."],
[".",".",".",".",".",".",".","."],
[".",".",".","B",".",".",".","."],
[".",".",".","p",".",".",".","."],
[".",".",".",".",".",".",".","."]]
输出：3

解题思路:

根据题目意思，白色的车分别沿着上下左右四个方向，当遇到白色的卒或者到达边界位置时停下来，否则遇到黑色的卒数量加一，然后停止往下走。

代码：

class Solution(object):
    def numRookCaptures(self, board):
        m = len(board)
        n = len(board[0])
        R_line ,R_column = 0, 0
        number = 0

        for i in range(m):
            for j in range(n):
                if board[i][j] == "R":
                    R_line = i
                    R_column = j
                    break

        for i in range(R_line-1,0,-1):       # 向上
            if board[i][R_column] != "B":
                if board[i][R_column] == "p":
                    number += 1
                    break
            else:
                break

        for i in range(R_line+1, m):       # 向下
            if board[i][R_column] != "B" :
                if board[i][R_column] == "p":
                    number += 1
                    break
            else:
                break

        for j in range(R_column-1, 0, -1):       # 向左
            if board[R_line][j] != "B" :
                if board[R_line][j] == "p":
                    number += 1
                    break
            else:
                break

        for j in range(R_column+1, n, 1):       # 向右
            if board[R_line][j] != "B":
                if board[R_line][j] == "p":
                    number += 1
                    break
            else:
                break

        return number

s = Solution()
board = [[".",".",".",".",".",".",".","."],
        [".",".",".","p",".",".",".","."],
        [".",".",".","R",".",".",".","p"],
        [".",".",".",".",".",".",".","."],
        [".",".",".",".",".",".",".","."],
        [".",".",".","p",".",".",".","."],
        [".",".",".",".",".",".",".","."],
        [".",".",".",".",".",".",".","."]]
print(s.numRookCaptures(board))

题目来源：
999. 车的可用捕获量

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