3 4 5 6 7 8 9 0 1 2 3 4 5 6 ***** from heapq import heappop, heapreplace, heapify from queue import PriorityQueue # Definition for singly-linked list. class ListNode(object): *** ListNode Class""" def_init__(self, val): self.val = val self.next = None
3 4 5 6 7 8 9 0 1 2 3 4 5 6 ***** from heapq import heappop, heapreplace, heapify from queue import PriorityQueue # Definition for singly-linked list. class ListNode(object): *** ListNode Class""" def_init__(self, val): self.val = val self.next = None
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Question
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Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
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from heapq import heappop, heapreplace, heapify
from queue import PriorityQueue
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11 class ListNode (object):
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""" ListNode Class"""
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36 def merge_k_lists (lists):
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""" Merge List """
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dummy = ListNode (None)
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# Definition for singly-linked list.
def __init__(self, val):
self.val = val
self.next = None
def merge_k_lists (lists):
""" Merge Lists """
dummy = node = ListNode (0)
list_h= [(n.val, n) for n in lists if n]
heapify(list_h)
while list_h:
_n_val = list_h[0]
if n_val.next is None:
heappop (list_h) #only change heap size when necessary
else:
heapreplace(list_h, (n_val.next.val, n_val.next))
node.next = n_val
node = node.next
return dummy.next
curr = dummy
q = PriorityQueue ()
for node in lists:
if node:
q.put ((node.val, node))
while not q.empty():
curr.next = q.get() [1] # These two lines seem to
curr = curr.next # be equivalent to :- curr = q.get() [1]
if curr.next:
q.put((curr.next.val, curr.next))
return dummy.next
I think my code's complexity is also 0(nlogk) and not using heap or priority queue,
n means the total elements and k means the size of list.
The merge TwoLists function in my code comes from the problem Merge Two Sorted Lists
whose complexity obviously is 0(n), n is the sum of length of 11 and 12.
To put it simpler, assume the k is 2^x, So the progress of combination is like a full binary tree,
from bottom to top. So on every level of tree, the combination complexity is n,
because every level have all n numbers without repetition.
The level of tree is x, ie log k. So the complexity is 0 (n log k).
for example, 8 ListNode, and the length of every ListNode is x1, x2,
x3, x4, x5, x6, x7, x8, total is n.
on level 3: x1+x2, x3+x4, x5+x6, x7+x8 sum: n
on level 2: x1+x2+x3+x4, x5+x6+x7+x8 sum: n
on level 1: x1+x2+x3+x4+x5+x6+x7+x8 sum: n](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F57bcf795-7b7b-45fd-a8ab-7fc3e4f54e2e%2F5bfa6f67-cd15-4666-86b1-e4ba9c630fee%2Fc8dlwo_processed.jpeg&w=3840&q=75)
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Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
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from heapq import heappop, heapreplace, heapify
from queue import PriorityQueue
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11 class ListNode (object):
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""" ListNode Class"""
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""" Merge List """
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dummy = ListNode (None)
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# Definition for singly-linked list.
def __init__(self, val):
self.val = val
self.next = None
def merge_k_lists (lists):
""" Merge Lists """
dummy = node = ListNode (0)
list_h= [(n.val, n) for n in lists if n]
heapify(list_h)
while list_h:
_n_val = list_h[0]
if n_val.next is None:
heappop (list_h) #only change heap size when necessary
else:
heapreplace(list_h, (n_val.next.val, n_val.next))
node.next = n_val
node = node.next
return dummy.next
curr = dummy
q = PriorityQueue ()
for node in lists:
if node:
q.put ((node.val, node))
while not q.empty():
curr.next = q.get() [1] # These two lines seem to
curr = curr.next # be equivalent to :- curr = q.get() [1]
if curr.next:
q.put((curr.next.val, curr.next))
return dummy.next
I think my code's complexity is also 0(nlogk) and not using heap or priority queue,
n means the total elements and k means the size of list.
The merge TwoLists function in my code comes from the problem Merge Two Sorted Lists
whose complexity obviously is 0(n), n is the sum of length of 11 and 12.
To put it simpler, assume the k is 2^x, So the progress of combination is like a full binary tree,
from bottom to top. So on every level of tree, the combination complexity is n,
because every level have all n numbers without repetition.
The level of tree is x, ie log k. So the complexity is 0 (n log k).
for example, 8 ListNode, and the length of every ListNode is x1, x2,
x3, x4, x5, x6, x7, x8, total is n.
on level 3: x1+x2, x3+x4, x5+x6, x7+x8 sum: n
on level 2: x1+x2+x3+x4, x5+x6+x7+x8 sum: n
on level 1: x1+x2+x3+x4+x5+x6+x7+x8 sum: n
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