合并具有限制的公共整数对

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号码并且公共号码位于第一个或最后一个位置(否则它已经全部连接)时,才会发生合并.

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.

又一个例子:

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

但它不适用于上述限制

Answer 1

下面的代码的工作原理如下：

1. 创建一个新列表，其大小为旧列表 + 1（输出可以达到的最大大小）。
2. 从对列表中的第一对开始，其第一个值位于新列表的末尾，第二个值位于新列表的开头。例如，使用 python 字典将它们标记为已访问。
3. 维护第一个和最后一个位置的两个索引，最初分别指向新列表的末尾和开头。
4. 迭代其余的对，如果对 i 的两个值在字典中存在或不存在，则跳过它。否则，将未访问过的值合并到第一个索引或最后一个索引处的元素（取决于访问过的值的位置），并更新索引。请注意，如果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 new list[第一个索引到 end] 与 new list[start 到最后一个索引] 的串联。

注意：每次合并操作都需要 O(1)。如果您知道对值的范围，您还可以使用布尔数组而不是字典。

#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]

该代码为您提到的所有测试用例提供了正确的输出。