python - 合并具有限制的公共(public)整数对

Question

我需要一种有效的方法来合并节点列表(整数对)。
只有当对中有一个公共(public)数字并且公共(public)数字在第一个或最后一个位置时才应该发生合并(否则它已经连接)。

例如:

mylist = [[4, 3], [6, 3]]
merge_links(mylist) # should output [4, 3, 6]

另一个例子:

mylist = [[4, 3], [6, 3], [6, 4]]
merge_links(mylist) 
# should output again [4, 3, 6] because both 6 and 4 allready exist in array.

还有一个例子:

mylist = [[4, 3], [6, 3], [6, 4], [6, 2], [7, 4], [4, 9]]
merge_links(mylist) 
# should output [7, 4, 3, 6, 2]

# [4, 3] ✔ 
# [4, 3] + [6, 3] ✔ -> [4, 3, 6]
# [4, 3, 6] + [6, 3] ✘ both 6 and 3 exist in [4, 3, 6] 
# [4, 3, 6] + [6, 4] ✘ both 6 and 4 exist in [4, 3, 6]
# [4, 3, 6] + [6, 2] ✔ -> [4, 3, 6, 2]
# [4, 3, 6, 2] + [7, 4] ✔ -> [7, 4, 3, 6, 2]
# [7, 4, 3, 6, 2] + [4, 9] ✘ 4 is allready connected "7-4-3"!

目前我正在使用:

def merge_links(a, b):
    inter = np.intersect1d(a, b, True)
    a = set(a) - set(inter)
    b = set(b) - set(inter)
    n = np.unique(list(a) + list(inter) + list(b))
    return n

但是对于上面的限制效果不是很好

Answer 1

下面的代码工作如下:

1. 创建一个新列表，其大小为旧列表的大小 + 1(最大大小 产量可达)。
2. 从你的对列表中的第一对开始，它的第一个值 在你的新列表的末尾，它的第二个值在 你的新 list 。并使用 python 字典将它们标记为已访问 例如。
3. 维护指向第一个和最后一个位置的两个索引 最初分别是新列表的结尾和开头。
4. 迭代其余的对，如果对 i 它的两个值都存在，或者 字典里没有，略过。否则，合并未访问的 第一个索引或最后一个索引处元素的值(取决于 访问值的位置)，并更新您的索引。注意 如果访问值不在第一个或 最后一个索引(如果它在中间某处)。
5. 返回新列表[第一个索引结束]与新的串联 列表[从最后一个索引开始]。

注意:每个合并操作都需要 O(1)。如果您知道对值的范围，也可以使用 bool 数组而不是字典。

#The function takes a list of pairs as an argument
def merge_lst(lst):

  #Dictionary to check if value is already in array
  visited = dict()

  #Length of old list
  lst_len = len(lst)

  #New list will have at most lst_len+1 elements
  new_lst = [0]*(lst_len+1)

  #Put the first pair values in last and first elements of the new list repectively and mark them as visited
  new_lst[lst_len], new_lst[0] = lst[0][0], lst[0][1]
  visited[lst[0][0]], visited[lst[0][1]] = True, True

  #Maintain the positions of your first and last elements, which are the the last index and 0 respectively now
  first_index, last_index = lst_len, 0

  #Iterate on the rest of pairs
  for i in range(1, lst_len):

    #Check if pair[a, b] are already visited
    a_exists, b_exists = lst[i][0] in visited, lst[i][1] in visited

    #Skip if they both exist or don't exist
    if(a_exists == b_exists):
      continue

    #Assume a was the common one
    common, to_merge = lst[i][0], lst[i][1]

    #If b exists (b is the common not a), then just swap
    if(b_exists):
      common, to_merge = lst[i][1], lst[i][0]

    #If the common element is at the first index, the first element and index are updated
    if(new_lst[first_index] == common):
      first_index-=1
      new_lst[first_index] = to_merge
      visited[to_merge] = True

    #If the common element is at the last index, the last element and index are updated
    elif(new_lst[last_index] == common):
      last_index+=1
      new_lst[last_index] = to_merge
      visited[to_merge] = True

    #Else, the common element is somewhre in the middle (already connected)

  #Return concatenation of new_lst[first_index to the end] with new_lst[0 to the last_index] 
  return new_lst[first_index:lst_len+1]+new_lst[0:last_index+1]

此代码为您提到的所有测试用例提供了正确的输出。