Question

我需要一种有效的方法来合并节点列表（成对的整数）。
只有在配对中有 一个公用号码 并且公用号码在第一个或最后一个位置是 时，才应进行合并（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"!

当前我正在使用：

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

但是对于上述限制来说效果不佳

Answer 1

下面的代码如下：

创建一个新列表，其大小为旧列表的大小+ 1（输出可以达到）。
从对列表中的第一对开始，并为其第一个值在新列表的末尾，第二个值在您的新列表。并使用python字典将其标记为已访问例如。
维护指向您的第一个和最后一个位置的两个索引首先分别是新列表的结尾和开头。
对其余对进行迭代，如果对i，则两个值都存在或字典中不存在，请跳过它。否则，合并未访问的对象元素在第一个索引或最后一个索引处的值（取决于访问值的位置），然后更新您的索引。注意如果访问的值不在第一个或第二个，则不会发生合并最后一个索引（如果它位于中间位置）。
用new返回新列表的连接[要结束的第一个索引] 列表[从最后一个索引开始]。

注意：每个合并操作都需要O（1）。如果知道成对值的范围，也可以使用布尔数组代替字典。

#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正确的输出。

有限制地合并公共整数对

1 个答案: