Question

我正在尝试解决Leetcode.com上的“ Available Captures for Rook”问题，但是我在尝试做的事情上遇到了一些困难（我是python的新手）。

因此，问题基本上是一个8x8的棋盘，其中有一个白鸦，几个白主教和许多黑典当，我必须计算白鸦可用的黑典当。首先，我获得了白色车的位置，然后获得了黑色棋子的所有位置，并将它们存储在称为“ positions”的元组列表中，然后我试图过滤那些与i或j不匹配的元素白车位置。

示例输入可能是：

[[".",".",".",".",".",".",".","."],
 [".",".",".","p",".",".",".","."],
 [".",".",".","R",".",".",".","p"],
 [".",".",".",".",".",".",".","."],
 [".",".",".",".",".",".",".","."],
 [".",".",".","p",".",".",".","."],
 [".",".",".",".",".",".",".","."],
 [".",".",".",".",".",".",".","."]]

和我的代码。

x = y = 0
count = 0
    positions = []
    for i in range(8):
        for j in range(8):
            if board[i][j] == "R":
                x = i
                y = j
            elif board[i][j] == "p":                    
                positions.append(tuple([i,j]))

    print(x,y)
    for pos in positions:
        if x != pos[0] and y != pos[1]:
            print(pos[0],pos[1])
            positions.remove(pos)
    return positions

所以我的职位列表原来是这样的：

[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), 
 (2, 1), (2, 2), (2, 4), (2, 5),
 (3, 1), (3, 5),
 (4, 1), (4, 2), (4, 4), (4, 5),
 (5, 1), (5, 2), (5, 3), (5, 4), (5, 5)]

，经过过滤后，我应该得到：

[(1, 3), (3, 1), (3, 5), (5, 3)]

但是我得到了：

[[1,2],[1,3],[1,5],[2,2],[2,5],[3,1],[3,5],[4,2],[4,5],[5,2],[5,3],[5,5]]

注意：这当然不是完整的解决方案，而是我尝试解决它的步骤。

比较元组列表python的单个值

0 个答案: