>假设我有 10x10 矩阵，其中包含以下数据：
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40
41 42 43 44 _ 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100
我的立场是[4][4].如何列出此位置的对角线值？
例如，预期结果将是：
[56, 67, 78, 89, 100, 1, 12, 23, 34]
[54, 63, 72, 81, 9, 18, 27, 36]
我当前的解决方案
def next?(index, row, size)
(((row + index) % size) + 1 ) % size
end
(1...chess_size).each do |l|
next_el, curr_el = next?(l, row, chess_size), (row + l) % chess_size
# this gets me the first diagnonal. Note that it prints out wrong value
tmp[0] << chess[curr_el][curr_el]
# this gets me the values from current row below to up
tmp[1] << chess[(row + l) % chess_size][row]
tmp[2] << chess[-l][l]
tmp[3] << chess[row][(row + l) % chess_size]
end
我们的矩阵将始终具有相同数量的行和列。
通常要从i和j中获取对角线值，您可以同时迭代i和j，直到其中一个为零。因此，主对角线(i-1, j-1), (i-2, j-2), ...到i, j >= 0和(i + 1, j + 1), (i +2, j + 2), ...i, j <= n.因为对角线(i - 1, j + 1), (i - 2, j + 2), ...可达i >= 0和j <= n，(i + 1, j-1), (i + 2, j - 2), ...可达i <= n和j >= 0。
这是破解者女王攻击问题的解决方案。
法典
def count_moves(n, obs, qrow, qcol)
qdiff = qrow-qcol
qsum = qrow+qcol
l = u = -1
r = d = n
ul = qdiff >= 0 ? qrow-qcol-1 : -1
dr = qdiff >= 0 ? n : qrow+n-qcol
ur = qsum < n ? -1 : qrow-n+qcol
dl = qsum < n ? qrow+qcol+1 : n
obs.uniq.each do |i,j|
case i <=> qrow
when -1 # up
case j <=> qcol
when -1 # up-left
ul = [ul,i].max
when 0 # up same col
u = [u,i].max
when 1 # up-right
ur = [ur,i].max
end
when 0 # same row
j < qcol ? (l = [l,j].max) : r = [r,j].min
else # down
case j <=> qcol
when -1 # down-left
dl = [dl,i].min
when 0 # down same col
d = [d,i].min
when 1 # down-right
dr = [dr,i].min
end
end
end
r + dl + d + dr - l - ul -u - ur - 8
end
例
假设棋盘有9行和列，女王的位置由字符q表示，每个障碍物都用字母o显示。所有其他位置由字母x表示。我们看到女王有16可能的动作(7上下，6左右，1左上到右对角线，2从右上到左下对角线。
arr = [
%w| x x x x x x x x x |, # 0
%w| o x x x x x x x x |, # 1
%w| x o x x x x x x x |, # 2
%w| x x o x x x x x o |, # 3
%w| x x x o x x x x x |, # 4
%w| x x x x x x o x x |, # 5
%w| o o x x x q x x x |, # 6
%w| x x x x x x o x x |, # 7
%w| x x x x x o x x x | # 8
# 0 1 2 3 4 5 6 7 8
]
qrow = qcol = nil
obs = []
n = arr.size
arr.each_with_index do |a,i|
a.each_with_index do |c,j|
case c
when 'o'
obs << [i,j]
when 'q'
qrow=i
qcol=j
end
end
end
qrow
#=> 6
qcol
#=> 5
obs
#=> [[1, 0], [2, 1], [3, 2], [3, 8], [4, 3], [5, 6], [6, 0], [6, 1], [7, 6], [8, 5]]
count_moves(n, obs, qrow, qcol)
#=> 16
解释
l是女王行中障碍物的最大列索引，小于女王的列索引;r是皇后区障碍物的斯莫尔斯柱索引大于皇后柱指数;u是女王列中小于女王行索引的障碍物的最大最大行索引;d是女王列中障碍物的最小行索引，大于女王的行索引;ul是女王左上角到右下对角线上小于女王行索引的障碍物的最大行索引;dr是女王左上角到右下对角线上障碍物的最小行索引，大于女王的行索引;ur是女王从右上角到左下对角线上小于女王行索引的障碍物的最大行索引;和dl是女王右上角到左下对角线上障碍物的最小行索引，大于女王的行索引。
对于上面的示例，在考虑障碍物之前，这些变量设置为以下值。
l = 0
r = 9
ul = 0
u = -1
ur = 2
dl = 9
d = 9
dr = 9
请注意，如果女王有行索引和列索引qrow和qcol，
i - j = qrow - qcol[i, j]女王左上至右下对角线上的所有位置;以及i + j = grow + gcol[i, j]女王右上角到左下角线的所有位置
然后，我们遍历所有(唯一)障碍物，确定每个障碍物是在女王行、女王列还是女王对角线之一中，然后用其行或列索引替换适用变量的值，如果它比以前最近的位置"更靠近"女王。
例如，如果障碍物位于女王行中，并且其列索引j小于女王的列索引，则进行以下计算：
l = [l, j].max
同样，如果障碍物位于女王的左上角到右下对角线上，并且其行索引i小于女王的行索引，则计算公式为：
ul = [ul, i].max
在考虑了上述示例中的所有障碍物后，变量具有以下值。
l #=> 1
r #=> 9
ul #=> 4
u #=> -1
ur #=> 5
dl #=> 9
d #=> 8
dr #=> 7
最后，我们计算女王可能移动到的方格总数。
qcol - l - 1 + # left
r - qcol - 1 + # right
u - qrow - 1 + # up
grow - d - 1 + # down
ul - qrow - 1 + # up-left
ur - qrow - 1 + # up-right
qrow - dl - 1 + # down-left
qrow - dr - 1 # down-right
简化为
r + dl + d + dr - l - ul -u - ur - 8
#=> 9 + 9 + 8 + 7 - 1 - 4 + 1 - 5 - 8 => 16
我已经应用了@OmG提供的逻辑。不确定它会有多高效。
def stackOverflow(matrixSize, *args)
pos, obstacles = *args
chess = (1..(matrixSize * matrixSize)).each_slice(matrixSize).to_a
obstacles.each do |l| chess[l[0]][l[1]] = '_' end
row, col = pos[:row] - 1, pos[:col] - 1
chess[row][col] = '♙'
directions = [[],[],[],[],[],[],[],[]]
(1...matrixSize).each do |l|
directions[0] << chess[row + l][col + l] if (row + l) < matrixSize && (col + l) < chess_size
directions[1] << chess[row - l][col - l] if (row - l) >= 0 && (col - l) >= 0
directions[2] << chess[row + l][col - l] if (row + l) < matrixSize && (col - l) >= 0
directions[3] << chess[row - l][col + l] if (row - l) >= 0 && (col + l) < matrixSize
directions[4] << chess[row + l][col] if row + l < matrixSize
directions[5] << chess[row - l][col] if row - l >= 0
directions[6] << chess[row][col + l] if col + l < matrixSize
directions[7] << chess[row][col - l] if col - l >= 0
end
end
stackOverflow(5, 3, {row: 4, col: 3}, [[4,4],[3,1],[1,2]] )
@CarySwoveland 似乎@Jamy正在解决黑客兰克queens-attack的另一个问题。
这个问题非常困难，因为这个想法是永远不要首先创建矩阵。也就是说，测试用例变得非常大，因此空间复杂性将成为一个问题。
我已经更改了我的实现，但由于超时问题仍然失败(这是因为测试用例变得非常大)。我不确定如何使其性能。
在我展示代码之前。让我用插图来解释我想要做什么：
这是我们的国际象棋：
---------------------------
| 1 2 3 4 5 |
| 6 7 8 9 10 |
| 11 12 13 14 15 |
| 16 17 18 19 20 |
| 21 22 23 24 25 |
---------------------------
这就是我们的女王所在的地方：queen[2][3]
---------------------------
| 1 2 3 4 5 |
| 6 7 8 9 10 |
| 11 12 13 ♙ 15 |
| 16 17 18 19 20 |
| 21 22 23 24 25 |
---------------------------
女王可以攻击所有8个方向。即：
horizontal(x2):
1. from queen position to left : [13, 12, 11]
2. from queen position to right : [15]
vertical(x2):
1. from queen position to top : [9, 4]
2. from queen position to bottom : [19, 24]
diagonal(x2):
1. from queen position to bottom-right : [20]
2. from queen position to top-left : [8, 2]
diagonal(x2):
1. from queen position to bottom-left : [18, 22]
2. from queen position to top-right : [10]
因为这8条路径内没有障碍物，女王一共可以攻击14次攻击。
假设我们有一些障碍：
---------------------------
| 1 2 3 4 5 |
| 6 7 x 9 10 |
| 11 x 13 ♙ 15 |
| 16 17 18 19 x |
| 21 x 23 x 25 |
---------------------------
现在女王总共可以攻击7次攻击：[13, 18, 19, 15, 10, 9, 4]
法典
MAXI = 10 ** 5
def queens_attack(size, number_of_obstacles, queen_pos, obstacles)
# exit the function if...
# size is negative or more than MAXI. Note MAXI has constraint shown in hackerrank
return if size < 0 || size > MAXI
# the obstacles is negative or more than the MAXI
return if number_of_obstacles < 0 || number_of_obstacles > MAXI
# the queen's position is outside of our chess dimension
return if queen_pos[:row] < 1 || queen_pos[:row] > size
return if queen_pos[:col] < 1 || queen_pos[:col] > size
# the queen's pos is the same as one of the obstacles
return if [[queen_pos[:row], queen_pos[:col]]] - obstacles == []
row, col = queen_pos[:row], queen_pos[:col]
# variable to increment how many places the queen can attack
attacks = 0
# the queen can attack on all directions:
# horizontals, verticals and both diagonals. So let us create pointers
# for each direction. Once the obstacle exists in the path, make the
# pointer[i] set to true
pointers = Array.new(8, false)
(1..size).lazy.each do |i|
# this is the diagonal from queen's pos to bottom-right
if row + i <= size && col + i <= size && !pointers[0]
# set it to true if there is no obstacle in the current [row + i, col + i]
pointers[0] = true unless [[row + i, col + i]] - obstacles != []
# now we know the queen can attack this pos
attacks += 1 unless pointers[0]
end
# this is diagonal from queen's pos to top-left
if row - i > 0 && col - i > 0 && !pointers[1]
# set it to true if there is no obstacle in the current [row - i, col - i]
pointers[1] = true unless [[row - i, col - i]] - obstacles != []
# now we know the queen can attack this pos
attacks += 1 unless pointers[1]
end
# this is diagonal from queen's pos to bottom-left
if row + i <= size && col - i > 0 && !pointers[2]
pointers[2] = true unless [[row + i, col - i]] - obstacles != []
attacks += 1 unless pointers[2]
end
# this is diagonal from queen's pos to top-right
if row - i > 0 && col + i <= size && !pointers[3]
pointers[3] = true unless [[row - i, col + i]] - obstacles != []
attacks += 1 unless pointers[3]
end
# this is verticle from queen's pos to bottom
if row + i <=size && !pointers[4]
pointers[4] = true unless [[row + i, col]] - obstacles != []
attacks += 1 unless pointers[4]
end
# this is verticle from queen's pos to top
if row - i > 0 && !pointers[5]
pointers[5] = true unless [[row - i, col]] - obstacles != []
attacks += 1 unless pointers[5]
end
# this is horizontal from queen's pos to right
if col + i <= size && !pointers[6]
pointers[6] = true unless [[row, col + i]] - obstacles != []
attacks += 1 unless pointers[6]
end
# this is horizontal from queen's pos to left
if col - i > 0 && !pointers[7]
pointers[7] = true unless [[row, col - i]] - obstacles != []
attacks += 1 unless pointers[7]
end
end
p attacks
end
问题
现在的问题是，我不知道为什么我的代码会从hackerrank中出现超时错误。我确实知道它，因为测试用例，国际象棋的维度可以是 10,000 X 10,000。但不知道我错过了什么约束。
我刚刚从 OP 发布的评论中了解到我解决了错误的问题，尽管 OP 的问题似乎很清楚，尤其是示例，并且与我的解释一致。我将把这个解决方案留给以下问题："给定一个数组arr，使得Matrix(*arr)是一个 NxM 矩阵，并且矩阵位置i,j，返回一个数组[d,a]，其中元素d和a是对角线和对角线上的元素，它们通过[d,a]但不包括[d,a]并且每个元素都旋转，以便第一个元素的行索引i+1如果i < arr.size-1，否则0。
法典
def diagonals(arr, row_idx, col_idx)
ncols = arr.first.size
sum_idx = row_idx+col_idx
diff_idx = row_idx-col_idx
a = Array.new(arr.size * arr.first.size) { |i| i.divmod(ncols) } -[[row_idx, col_idx]]
[a.select { |r,c| r-c == diff_idx }, a.select { |r,c| r+c == sum_idx }].
map do |b| b.sort_by { |r,_| [r > row_idx ? 0:1 , r] }.
map { |r,c| arr[r][c] }
end
end
数组的所有元素arr大小必须相等，但没有要求arr.size = arr.first.size.
例
arr = [
[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
[11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
[21, 22, 23, 24, 25, 26, 27, 28, 29, 30],
[31, 32, 33, 34, 35, 36, 37, 38, 39, 40],
[41, 42, 43, 44, 45, 46, 47, 48, 49, 50],
[51, 52, 53, 54, 55, 56, 57, 58, 59, 60],
[61, 62, 63, 64, 65, 66, 67, 68, 69, 70],
[71, 72, 73, 74, 75, 76, 77, 78, 79, 80],
[81, 82, 83, 84, 85, 86, 87, 88, 89, 90],
[91, 92, 93, 94, 95, 96, 97, 98, 99, 100]
]
diagonals(arr, 4, 4)
#=> [[56, 67, 78, 89, 100, 1, 12, 23, 34],
# [54, 63, 72, 81, 9, 18, 27, 36]]
解释
假设
arr = (1..16).each_slice(4).to_a
#=> [[ 1, 2, 3, 4],
# [ 5, 6, 7, 8],
# [ 9, 10, 11, 12],
# [13, 14, 15, 16]]
row_idx = 2
col_idx = 1
步骤如下。
a = Array.new(arr.size) { |i| Array.new(arr.first.size) { |j| [i,j] } }
#=> [[[0, 0], [0, 1], [0, 2], [0, 3]],
# [[1, 0], [1, 1], [1, 2], [1, 3]],
# [[2, 0], [2, 1], [2, 2], [2, 3]],
# [[3, 0], [3, 1], [3, 2], [3, 3]]]
ncols = arr.first.size
#=> 4
sum_idx = row_idx+col_idx
#=> 3
diff_idx = row_idx-col_idx
#=> 1
a = Array.new(arr.size * arr.first.size) { |i| i.divmod(ncols) } - [[row_idx, col_idx]]
#=> [[0, 0], [0, 1], [0, 2], [0, 3], [1, 0], [1, 1], [1, 2], [1, 3],
# [2, 0], [2, 2], [2, 3], [3, 0], [3, 1], [3, 2], [3, 3]]
选择并排序[r, c]通过[row_idx, col_idx]的左上角到右下对角线上的位置。
b = a.select { |r,c| r-c == diff_idx }
#=> [[1, 0], [3, 2]]
c = b.sort_by { |r,_| [r > row_idx ? 0:1 , r] }
#=> [[3, 2], [1, 0]]
选择并排序[r, c]通过[row_idx, col_idx]的右下角对角线上的位置。
d = a.select { |r,c| r+c == sum_idx }
#=> [[0, 3], [1, 2], [3, 0]]
e = d.sort_by { |r,c| [r > row_idx ? 0:1 , r] }
#=> [[3, 0], [0, 3], [1, 2]]
[c, e].map { |f| f.map { |r,c| arr[r][c] }
#=> [c, e].map { |f| f.map { |r,c| arr[r][c] } }
#=> [[15, 5], [13, 4, 7]]
我刚刚从 OP 发布的评论中了解到我解决了错误的问题，尽管 OP 的问题似乎很清楚，尤其是示例，并且与我的解释一致。我将把这个解决方案留给以下问题："给定一个数组arr，使得Matrix(*arr)是一个 NxM 矩阵，并且矩阵位置i,j，返回一个数组[d,a]，其中元素d和a是对角线和反对角线上的元素，它们通过[d,a]但不包括[d,a]并且每个都旋转，以便第一个元素的行索引i+1如果i < arr.size-1，否则0。
以下方法使用 Matrix 类中的方法。
法典
require 'matrix'
def diagonals(arr, row_idx, col_idx)
[diag(arr, row_idx, col_idx),
diag(arr.map(&:reverse).transpose, arr.first.size-1-col_idx, row_idx)]
end
def diag(arr, row_idx, col_idx)
nrows, ncols = arr.size, arr.first.size
lr = [ncols-col_idx, nrows-row_idx].min - 1
ul = [col_idx, row_idx].min
m = Matrix[*arr]
[*m.minor(row_idx+1, lr, col_idx+1, lr).each(:diagonal).to_a,
*m.minor(row_idx-ul, ul, col_idx-ul, ul).each(:diagonal).to_a]
end
例
arr = [
[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
[11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
[21, 22, 23, 24, 25, 26, 27, 28, 29, 30],
[31, 32, 33, 34, 35, 36, 37, 38, 39, 40],
[41, 42, 43, 44, 45, 46, 47, 48, 49, 50],
[51, 52, 53, 54, 55, 56, 57, 58, 59, 60],
[61, 62, 63, 64, 65, 66, 67, 68, 69, 70],
[71, 72, 73, 74, 75, 76, 77, 78, 79, 80],
[81, 82, 83, 84, 85, 86, 87, 88, 89, 90],
[91, 92, 93, 94, 95, 96, 97, 98, 99, 100]
]
diagonals arr, 4, 4
#=> [[56, 67, 78, 89, 100, 1, 12, 23, 34], [54, 63, 72, 81, 9, 18, 27, 36]]
diagonals arr, 4, 5
#=> [[57, 68, 79, 90, 2, 13, 24, 35], [55, 64, 73, 82, 91, 10, 19, 28, 37]]
diagonals arr, 0, 9
#=> [[], [19, 28, 37, 46, 55, 64, 73, 82, 91]]
解释
假设数组和目标位置如下所示。
arr = (1..30).each_slice(6).to_a
#=> [[ 1, 2, 3, 4, 5, 6],
# [ 7, 8, 9, 10, 11, 12],
# [13, 14, 15, 16, 17, 18],
# [19, 20, 21, 22, 23, 24],
# [25, 26, 27, 28, 29, 30]]
row_idx = 2
col_idx = 3
注arr[2][3] #=> 16.我们通过计算两个矩阵次要矩阵的对角线来获得负斜率的对角线：
[[23, 24],
[29, 30]]
和
[[2, 3],
[8, 9]]
给我们
[*[23, 30], *[2, 9]]
#=> [23, 30, 2, 9]
为了获得另一个对角线，我们将数组逆时针旋转 90 度，调整row_idx和col_idx并重复上述过程。
arr.map(&:reverse).transpose
#=> [[6, 12, 18, 24, 30],
# [5, 11, 17, 23, 29],
# [4, 10, 16, 22, 28],
# [3, 9, 15, 21, 27],
# [2, 8, 14, 20, 26],
# [1, 7, 13, 19, 25]]
ncols = arr.first.size
#=> 6
row_idx, col_idx = ncols-1-col_idx, row_idx
#=> [2, 2]
我们现在从矩阵次要元素中提取对角线
[[21, 27],
[20, 26]]
和
[[6, 12],
[5, 11]]
要获得第二个对角线：
[21, 26, 6, 11]
def possible_moves(val):
# val is a value between 0 and n*n-1
for i in range(n*n):
if i == val:
board[i // n][i % n] = 'Q'
continue
#mark row and column with a dot
if i % n == val % n or i // n == val // n:
board[i//n][i%n] = '.'
# mark diagonals with a dot
if i % (n + 1) == val % (n + 1) and abs(i % n - val % n) == abs(i // n - val // n):
board[i//n][i%n] = '.'
if i % (n - 1) == val % (n - 1) and abs(i % n - val % n) == abs(i // n - val // n):
board[i//n][i%n] = '.'
n = 10 #board size = n x n
board = [['0' for x in range(n)] for y in range(n)] #initialize board with '0' in every row and col
possible_moves(40)
0"和她可以移动的".">