同步两个数组[重复]

在这里实现 TNT's elegant solution 用于删除数组的第一个匹配元素

class Array
  def delete_first item
    delete_at(index(item) || length)
  end
  def distinct other, own = self.dup
    other.each{|e| own.delete_first(e)}
    own
  end
end

arr1.distinct arr2 # ["a", "a", "f"]
arr2.distinct arr1 # ["b", "d"]

不漂亮!

def array_compare(arr1, arr2)
  arr2 = arr2.clone
  add_to = []
  arr1.each do |el|
    if index = arr2.index(el)
      arr2.delete_at(index)
    else
      add_to << el
    end
  end
  return add_to
end

-> arr1 = ['a', 'a', 'a', 'b', 'c,', 'f']
["a", "a", "a", "b", "c,", "f"]

-> arr2 = ['a', 'b', 'b', 'c,', 'd']
["a", "b", "b", "c,", "d"]

-> array_compare(arr1, arr2)
["a", "a", "f"]

-> array_compare(arr2, arr1)
["b", "d"]

1. 计算每个数组中的所有出现次数,并存储在两个散列中
3. 为哈希1中的每个附加计数输出数组项
4. 扁平化和输出

h1 = arr1.inject(Hash.new(0)) { |total, e| total[e] += 1 ;total }
# => {"a"=>3, "b"=>1, "c"=>1, "f"=>1
h2 = arr2.inject(Hash.new(0)) { |total, e| total[e] += 1 ;total }
# => {"a"=>1, "b"=>2, "c"=>1, "d"=>1}

h1.map { |k, v| [k] * [v - h2[k], 0].max }.flatten
# => ["a", "a", "f"]
h2.map { |k, v| [k] * [v - h1[k], 0].max }.flatten
# => ["b", "d"]

当允许重复时,Ruby的常规数组算法几乎没有用处。 尝试逐个删除元素:

# Original example
arr1 = ['a', 'a', 'a', 'b', 'c', 'f']
arr2 = ['a', 'b', 'b', 'c', 'd']

# Removing elements from arr1 that exist in arr2 ("arr1 - arr2")
arr1wo2 = arr1.dup
arr2.each { |val| arr1wo2.delete_at(arr1wo2.index(val) || arr1wo2.length) }
arr1wo2  # => ["a", "a", "f"]

# Removing elements from arr2 that exist in arr1 ("arr2 - arr1")
arr2wo1 = arr2.dup
arr1.each { |val| arr2wo1.delete_at(arr2wo1.index(val) || arr2wo1.length) }
arr2wo1  # => ["b", "d"]

希望你能帮上忙。