在二维表检查对角线（蟒蛇） [英] checking diagonals in 2d list (Python)

问题描述

最初的问题：对于给定的3x3的井字板检查，如果一个玩家赢得了

我来了迄今最简单的办法是旋转矩阵，总结各行：

 板
[0，1，2]，[3，4，5〕，〔6，7，8]]

PR（板）
0 1 2
3 4 5
6 7 8

PR（邮政编码（*板））
0 3 6
1 4 7
2 5 8
 

0..9数字上面只为表现出对董事会职位，通常他们会装满1，供玩家1和-1的球员2和0的空缺职位。去逐行，如果它总结为3或-3，这是获胜的块。

不过，对角线不检查。有没有一些方法来提取这种矩阵对角线优雅+具有高性能的方式？我的意思不是用平凡索引手动（0，1，2），而是获得对角n阶矩阵。

P.S。 PR只是一个辅助功能，打印二维列表：

 高清PR（X）：
    对于排在X：
        打印''。加入（图（STR，行））
 

解决方案

与数幻方的游戏领域

  2 | 9 | 4
7 | 5 | 3
6 | 1 | 8
 

现在经过三年移动总结和检查总和是15 - >赢家。你要检查这个每一个球员。当然，你有4招和第5手（只有开始游戏的玩家）

后复查

这是我在我的第一个Java类解决了这个问题。

The initial problem: for a given 3x3 tic tac toe board check if one of the players has won.

The simplest solution I have come up so far is rotating the matrix and summing up each row:

board
[[0, 1, 2], [3, 4, 5], [6, 7, 8]]

pr(board)
0 1 2
3 4 5
6 7 8

pr(zip(*board))
0 3 6
1 4 7
2 5 8

0..9 numbers above are just for showing positions on the board, normally they would be filled with 1 for player 1 and -1 for player 2 and 0 for unfilled position. Go row by row and if it sums up to 3 or -3 this is the winning block.

However, diagonals are not checked. Is there some way to extract diagonals from such matrix in elegant + highly performant way? I don't mean using trivial indexes "manually" (0, 1, 2), but rather get diagonal of n x n matrix.

P.S. pr is just a helper function for printing 2d list:

def pr(x):
    for row in x:
        print ' '.join(map(str, row))

解决方案

Number your game field with a magic square

2|9|4
7|5|3
6|1|8

Now sum up after three moves and check if the sum is 15 --> Winner. You have to check this for every player. Surely you have to recheck after 4th move and 5th move (only the player that started the game)

This is how I solved this problem in my first Java class.

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