4

获得两个数组的最长公共前缀（从原始索引 0 开始的子数组）和后缀（以原始索引 -1 结尾的子数组）的最佳方法是什么？例如，给定两个数组：

[:foo, 1, :foo, 0, nil, :bar, "baz", false]
[:foo, 1, :foo, 0, true, :bar, false]

这些数组的最长公共前缀是：

[:foo, 1, :foo, 0]

这些数组的最长公共后缀是：

[false]

当原始数组中索引 0/-1 处的元素不同时，公共前缀/后缀应该是一个空数组。

4

5 回答 5

3

一种可能的解决方案：

a1 = [:foo, 1, 0, nil, :bar, "baz", false]
a2 = [:foo, 1, 0, true, :bar, false]

a1.zip(a2).take_while { |x, y| x == y }.map(&:first)
#=> [:foo, 1, 0]

反转输入数组和输出数组会发现一个共同的后缀：

a1.reverse.zip(a2.reverse).take_while { |x, y| x == y }.map(&:first).reverse
#=> [false]

边缘情况：zip用值填充“参数”数组nil：

a1 = [true, nil, nil]
a2 = [true]

a1.zip(a2).take_while { |x, y| x == y }.map(&:first)
#=> [true, nil, nil]

这可以通过将初始数组截断为第二个数组的长度来避免：

a1[0...a2.size].zip(a2).take_while { |x, y| x == y }.map(&:first)
于 2014-08-20T13:47:00.733 回答
2

另一种没有边缘情况的解决方案：

a1 = [:foo, 1, 0, nil, :bar, "baz", false]
a2 = [:foo, 1, 0, true, :bar, false]

a1.take_while.with_index {|v,i| a2.size > i && v == a2[i] }

编辑：提高性能

class Array
  def common_prefix_with(other)
    other_size = other.size
    take_while.with_index {|v,i| other_size > i && v == other[i] }
  end
end

a1.common_prefix_with a2
于 2014-08-20T14:45:45.503 回答
1

三种解决方案：粗野（#3，初始答案）、更好（#2，第一次编辑）和最佳（#1，@Stefan 答案的变体，第二次编辑）。

a = [:foo, 1, :foo, 0, nil, :bar, "baz", false]
b = [:foo, 1, :foo, 0, true, "baz", false]

c = [:foo,1,:goo]
d = [:goo,1,:new]

注意b与 OP 的示例略有不同。

除非下面另有说明，否则将通过反转数组、应用common_prefix然后反转结果来计算公共后缀。

#1

Stefan 的答案的变体，它摆脱了zip并且map（并保留了他将一个数组截断为最多另一个数组长度的技术）：

def common_prefix(a,b)
  a[0,b.size].take_while.with_index { |e,i| e == b[i] }
end

common_prefix(a,b)
  #=> [:foo, 1, :foo, 0]
common_prefix(c,d)
  #=> []

#2

def common_prefix(a,b)
  any, arr = a.zip(b).chunk { |e,f| e==f }.first
  any ? arr.map(&:first) : []
end

def common_suffix(a,b)
  any, arr = a[a.size-b.size..-1].zip(b).chunk { |e,f| e==f }.to_a.last
  any ? arr.map(&:first) : []
end

common_prefix(a,b)
  #=> [:foo, 1, :foo, 0]
  # Nore: any, arr = a.zip(b).chunk { |e,f| e==f }.to_a
  #  => [[true, [[:foo, :foo], [1, 1], [:foo, :foo], [0, 0]]],
  #      [false, [[nil, true], [:bar, :baz], ["baz", false], [false, nil]]]]

common_suffix(a,b)
  #=> ["baz", false]
  # Note: any, arr = a[a.size-b.size..-1].zip(b).chunk { |e,f| e==f }.to_a
  #  => [[false, [[1, :foo], [:foo, 1], [0, :foo], [nil, 0], [:bar, true]]],
  #      [true, [["baz", "baz"], [false, false]]]]

当:first被发送到枚举器Enumerable#chunk时，枚举器的第一个元素被返回。因此，它的效率应该与使用Enumerable#take_while 相当。

common_prefix(c,d)
  #=> []
common_suffix(c,d)
  #=> []

#3

def common_prefix(a,b)
  a[0,(0..[a.size, b.size].min).max_by { |n| (a[0,n]==b[0,n]) ? n : -1 }]
end

common_prefix(a,b)
  #=> [:foo, 1, :foo, 0]

common_prefix(c,d)
  #=> []
于 2014-08-20T15:26:48.067 回答
1

Gentlemen, start your engines!

Methods tested

I tested the second method given by @stefan, both methods proposed by @BroiSatse and the three methods I offered.

class Array #for broiSatse2
  def common_prefix_with(other)
    other_size = other.size
    take_while.with_index {|v,i| other_size > i && v == other[i] }
  end
end

class Cars
  def self.stefan(a1,a2)
    a1[0...a2.size].zip(a2).take_while { |x, y| x == y }.map(&:first)
  end

  def self.broiSatse1(a1,a2)
    a1.take_while.with_index {|v,i| a2.size > i && v == a2[i] }
  end

  def self.broiSatse2(a1,a2)
    a1.common_prefix_with(a2)
  end

  def self.cary1(a,b)
    a[0,b.size].take_while.with_index { |e,i| e == b[i] }
  end

  def self.cary2(a,b)
    any, arr = a.zip(b).chunk { |e,f| e==f }.first
    any ? arr.map(&:first) : []
  end

  def self.cary3(a,b)
    a[0,(0..[a.size, b.size].min).max_by { |n| (a[0,n]==b[0,n]) ? n : -1 }]
  end
end

I did not include the solution given by @6ftDan (not to be confused with 5' Dan or 7' Dan) because it did not pass all the tests.

Construction of test arrays

random_arrays(n) constructs one pair of arrays. n is the size of the smaller of the two. The larger is of size n+1.

def random_arrays(n)
  m = rand(n)
  # make arrays the same for the first m elements
  a = Array.new(m,0)
  b = Array.new(m,0)
  if m < n
    # make the m+1 elements different
    a << 0
    b << 1
    # randomly assign 0s and 1a to the remaining elements
    (n-m-1).times { a << rand(2); b << rand(2) }  if m < n - 1
  end
  # make arrays unequal in length
  (rand(2) == 0) ? a << 0 : b << 0
  [a,b]
end

Confirm methods tested give identical results

N = 10000 #size of smaller of two input arrays
methods = Cars.methods(false)

# ensure are methods produce the same results and that
# all deal with edge cases properly
20.times do |i|
  test = case i
         when 0 then [[0,1],[1,1]]
         when 1 then [[0],[]]
         when 1 then [[0,0,0],[0,0]]
         else
         random_arrays(N)
         end  
  soln = Cars.send(methods.first, *test)
  methods[1..-1].each  do |m|
    unless soln == Cars.send(m, *test)
      puts "#{m} has incorrect solution for #{test}"
      exit
    end
  end
end

puts "All methods yield the same answers\n"

Benchmarking

require 'benchmark'

I = 1000 #number of array pairs to test

@arr = I.times.with_object([]) { |_,a| a << random_arrays(N) } #test arrays

#test method m 
def testit(m)
  I.times { |i| Cars.send(m, *@arr[i]) }
end    

Benchmark.bm(12) { |bm| methods.each { |m| bm.report(m) { testit(m) } } end

Results

All methods yield the same answers

                   user     system      total        real
stefan        11.260000   0.050000  11.310000 ( 11.351626)
broiSatse1     0.860000   0.000000   0.860000 (  0.872256)
broiSatse2     0.720000   0.010000   0.730000 (  0.717797)
cary1          0.680000   0.000000   0.680000 (  0.684847)
cary2         13.130000   0.040000  13.170000 ( 13.215581)
cary3         51.840000   0.120000  51.960000 ( 52.188477)
于 2014-08-21T03:02:13.510 回答
0

这适用于独特的数组。

字首

x = [:foo, 1, 0, nil, :bar, "baz", false]
y = [:foo, 1, 0, true, :bar, false]
(x & y).select.with_index {|item,index| x.index(item) == index}

输出： => [:foo, 1, 0]

在 Arrays 上反向运行以获取 Affix

(x.reverse & y.reverse).select.with_index {|item,index| x.reverse.index(item) == index}

输出：=> [false]

于 2014-08-20T13:38:51.457 回答