代码

Question

说我有一个二维数组，像这样：

Emaill.setText("test");

如何使用上面这些Arros的交集或并集来获得以下结果：

[ [0, 1], [2, 3], [0, 4] ]

解释以上内容：

使用[[0, 1, 4], [2, 3]]的原因是因为0被连接到1和4
[0, 1, 4]的原因是因为2仅连接到3

我们如何使用集合相交或并集来做到这一点？我很可能。

代码

我当前的实现实际上是在创建[2,3]并寻找邻居：

Node

上面的实现将输出预期的结果，但对于大多数其他情况则不会。

更新

对于案例def connected_neighbors(astronaut) graph, to_return, node_a, node_b = {}, [], nil, nil astronaut.each do |city_a, city_b| node_a, node_b = (graph[city_a] || Node.new(city_a)), (graph[city_b] || Node.new(city_b)) node_a.connect node_b graph[city_a] = node_a unless graph[city_a] end graph.each do |key,_| node = graph[key] to_return << [node.key, node.neighbors.collect(&:key)].flatten end to_return end

输出应为[1, 2], [2, 3]

这是因为数组中的范围是从0到3。

所以因为数组中不存在0，所以它将是单独的

Answer 1

这也许不是形成不相交数组的最有效方法，但是它确实产生了预期的结果。可以很容易地通过矛盾确定它起作用的证据。

arr = [[0,2], [1,3], [4,6], [7,9], [6,8], [5,7], [2,4], [3,7], [10,11]]

require 'set'

sets = arr.map(&:to_set)
  #=> [#<Set: {0, 2}>, #<Set: {1, 3}>, #<Set: {4, 6}>, #<Set: {7, 9}>, #<Set: {6, 8}>,
  #    #<Set: {5, 7}>, #<Set: {2, 4}>, #<Set: {3, 7}>, #<Set: {10, 11}>]

loop do
  break if sets.size == 1
  set1, set2 = sets.combination(2).find { |set1,set2| (set1 & set2).any? }
  break if set1.nil?
  set1.replace(set1 | set2)
  sets.delete(set2)
end

sets.map(&:to_a)
  #=> [[0, 2, 4, 6, 8], [1, 3, 7, 9, 5], [10, 11]]

我使用集合而不是数组来加快联合和相交的计算。

可以通过包含一些puts语句来说明这些步骤。

sets = arr.map(&:to_set)

loop do
  puts "(#{sets.size} sets at beginning of loop"
  puts "  #{sets}"
  puts "  break as sets.size == 1" if sets.size == 1
  break if sets.size == 1
  set1, set2 = sets.combination(2).find { |set1,set2| (set1 & set2).any? }
  if set1.nil?
    puts "    After find, set1 = nil, so break" if set1.nil?
  else
    puts "    After find, set1 = #{set1}"
    puts "                set2 = #{set2}"
  end
  break if set1.nil?
  set1.replace(set1 | set2)
  sets.delete(set2)
  puts "  sets after set1 |= set2 and sets.delete(set2)"
  puts "  #{sets}" 
end

sets.map(&:to_a)

打印以下内容。

(9) sets at beginning of loop
  [#<Set: {0, 2}>, #<Set: {1, 3}>, #<Set: {4, 6}>, #<Set: {7, 9}>, #<Set: {6, 8}>,
   #<Set: {5, 7}>, #<Set: {2, 4}>, #<Set: {3, 7}>, #<Set: {10, 11}>]
    After find, set1 = #<Set: {0, 2}>
                set2 = #<Set: {2, 4}>
 sets after set1 |= set2 and sets.delete(set2)
   [#<Set: {0, 2, 4}>, #<Set: {1, 3}>, #<Set: {4, 6}>, #<Set: {7, 9}>,
    #<Set: {6, 8}>, #<Set: {5, 7}>, #<Set: {3, 7}>, #<Set: {10, 11}>]

(8) sets at beginning of loop
  [#<Set: {0, 2, 4}>, #<Set: {1, 3}>, #<Set: {4, 6}>, #<Set: {7, 9}>,
   #<Set: {6, 8}>, #<Set: {5, 7}>, #<Set: {3, 7}>, #<Set: {10, 11}>]
    After find, set1 = #<Set: {0, 2, 4}>
                set2 = #<Set: {4, 6}>
 sets after set1 |= set2 and sets.delete(set2)
   [#<Set: {0, 2, 4, 6}>, #<Set: {1, 3}>, #<Set: {7, 9}>, #<Set: {6, 8}>,
    #<Set: {5, 7}>, #<Set: {3, 7}>, #<Set: {10, 11}>]

(7) sets at beginning of loop
  [#<Set: {0, 2, 4, 6}>, #<Set: {1, 3}>, #<Set: {7, 9}>, #<Set: {6, 8}>,
   #<Set: {5, 7}>, #<Set: {3, 7}>, #<Set: {10, 11}>]
    After find, set1 = #<Set: {0, 2, 4, 6}>
                set2 = #<Set: {6, 8}>
 sets after set1 |= set2 and sets.delete(set2)
   [#<Set: {0, 2, 4, 6, 8}>, #<Set: {1, 3}>, #<Set: {7, 9}>, #<Set: {5, 7}>,
    #<Set: {3, 7}>, #<Set: {10, 11}>]

(6) sets at beginning of loop
  [#<Set: {0, 2, 4, 6, 8}>, #<Set: {1, 3}>, #<Set: {7, 9}>, #<Set: {5, 7}>,
   #<Set: {3, 7}>, #<Set: {10, 11}>]
    After find, set1 = #<Set: {1, 3}>
                set2 = #<Set: {3, 7}>
 sets after set1 |= set2 and sets.delete(set2)
   [#<Set: {0, 2, 4, 6, 8}>, #<Set: {1, 3, 7}>, #<Set: {7, 9}>, #<Set: {5, 7}>,
    #<Set: {10, 11}>]

(5) sets at beginning of loop
  [#<Set: {0, 2, 4, 6, 8}>, #<Set: {1, 3, 7}>, #<Set: {7, 9}>, #<Set: {5, 7}>,
   #<Set: {10, 11}>]
    After find, set1 = #<Set: {1, 3, 7}>
                set2 = #<Set: {7, 9}>
 sets after set1 |= set2 and sets.delete(set2)
   [#<Set: {0, 2, 4, 6, 8}>, #<Set: {1, 3, 7, 9}>, #<Set: {5, 7}>, #<Set: {10, 11}>]

(4) sets at beginning of loop
  [#<Set: {0, 2, 4, 6, 8}>, #<Set: {1, 3, 7, 9}>, #<Set: {5, 7}>, #<Set: {10, 11}>]
    After find, set1 = #<Set: {1, 3, 7, 9}>
                set2 = #<Set: {5, 7}>
 sets after set1 |= set2 and sets.delete(set2)
   [#<Set: {0, 2, 4, 6, 8}>, #<Set: {1, 3, 7, 9, 5}>, #<Set: {10, 11}>]

(3) sets at beginning of loop
  [#<Set: {0, 2, 4, 6, 8}>, #<Set: {1, 3, 7, 9, 5}>, #<Set: {10, 11}>]
    After find, set1 = nil, so break

sets.map(&:to_a)
  #=> [[0, 2, 4, 6, 8], [1, 3, 7, 9, 5], [10, 11]]

Answer 2

如果我了解您的要求，也许您可以分组：

astr = [ [0, 1], [1, 3], [3, 4], [2, 5], [5, 6] ]

mapp = astr.map.with_index do |_, i|
  res = []
  astr[i..-1].each do |e|
    if res.empty?
      res = res && e 
    else
      res = res + e unless (res & e).empty?
    end
  end
  res.uniq
end.slice_when { |j, k| j.size <= k.size }.map(&:first)


mapp #=> [[0, 1, 3, 4], [2, 5, 6]]

对于astr = [ [0, 1], [1, 3], [3, 4], [2, 5], [5, 0] ]，它返回

#=> [[0, 1, 3, 4, 5], [2, 5, 0]]

对于astr = [ [1, 2], [2, 3] ]，它返回

#=> [[1, 2, 3]]

如果结果大小为1以下，请随时使用.unshift [0]。

如何在ruby中的2D数组上使用设置交集或并集？

代码

更新

2 个答案: