999.Available Capture for rook

999.Available Captures for Rook

 

车的可用捕获量

 

在一个8 × 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。</br>

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。<br>返回车能够在一次移动中捕获到的卒的数量。

 

示例：

 

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]

 

输出：3

 

解释：在本例中，车能够捕获所有的卒。

 

Solution 1:

def numRockCaptures(board):
    """
    :type board: List[List[str]]
    :rtype: int
    """
    result = 0
    index = 0
    for i in board:
        if 'R' in i:
            row = ''.join(z for z in i if z != '.')
            if 'Rp' in row:
                result += 1
            if 'pR' in row:
                result += 1
            index = i.index("R")
            break
    col = ''.join(board[i][index] for i in range(8) if board[i][index] != '.')
    if 'Rp' in col:
        result += 1
    if 'pR' in col: 
        result += 1

    return result

if __name__ == '__main__':
    """
    找到车的位置，然后将车所在的list转为str
    判断str中是否存在'Rp'和'pR'
    然后再获取车所在列的list，将它转为str
    判断str中是否存在'Rp'和'pR'
    """
    # test case
    board = [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
    result = numRockCaptures(board)
    print(result)

Solution 2:

def numRockCaptures(board):
    """
    :type board: List[List(str)]
    :reype: int
    """
    for i in range(8):
        for j in range(8):
            if board[i][j] == 'R':
                x0, y0 = i, j
    
    res = 0
    for i, j in [[1, 0], [0, 1], [-1, 0], [0, -1]]:
        x, y = x0 + i, y0 + j
        while 0 <= x < 8 and 0 <= y < 8:
            if board[x][y] == 'p':
                res += 1
                break
            if board[x][y] != '.':
                break
            x, y = x + i, y + j
    return res

if __name__ == '__main__':
    """
    先找到车R的位置
    然后依次从四个方向去遍历
    """
    # test case
    board = [[".",".",".",".",".",".",".","."],[".","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",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
    print(numRockCaptures(board))