具有限制的公共整数上的合并对

Question

我需要一种有效的方法来合并一个节点列表(整数对)。

只有当对中有一个公共数字，并且公共数字是位于第一个或最后一个位置时，合并才会发生(否则它就连接起来了)。

例如：

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

另一个例子是：

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.

还有另一个例子：

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，对于上述限制，它不起作用。

Answer 1

以下代码的工作原理如下：

1. 创建一个大小为旧列表+1的新列表(输出可以达到的最大大小)。
2. 从列表中的第一对开始，它的第一个值在新列表的末尾，第二个值在新列表的开始。并使用python字典将它们标记为已访问的。
3. 维护第一个和最后一个位置的两个索引，分别指向新列表的末尾和开始位置。
4. 对其余的对进行迭代，如果对于对i，它的两个值在字典中都存在或不存在，则跳过它。否则，将未访问的值合并为第一个索引或最后一个索引的元素(取决于访问值的位置)，并更新索引。注意，如果所访问的值不在第一个或最后一个索引(如果它位于中间的某个位置)，则不会发生合并。
5. 返回新listfirst索引的连接，以新liststart结束到最后一个索引。

注意:每个合并操作都采用O(1)。此外，如果您知道对值的范围，则可以使用布尔值数组而不是字典。

#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。