[英]reverse print an immutable linked list with less than O(n) space
解決這個問題,我的想法是遞歸的,在每次遞歸過程中,反向打印鏈表的下半部分,然后反向打印鏈表的上半部分。 因此額外的空間是O(log n)
- 這是用於遞歸堆棧的額外空間,但是它超過O(n)的時間(O(n log n) - 在每個(log n)級別上的組合調用遞歸迭代整個列表將每個部分切成兩半)。
是否存在實現相同目標的算法 - 反向打印具有少於O(n)空間且最多為O(n)時間的不可變單鏈表?
源代碼 (Python 2.7):
class LinkedListNode:
def __init__(self, value, next_node):
self.value = value
self.next_node = next_node
@staticmethod
def reverse_print(list_head, list_tail):
if not list_head:
return
if not list_head.next_node:
print list_head.value
return
if list_head == list_tail:
print list_head.value
return
p0 = list_head
p1 = list_head
while p1.next_node != list_tail and p1.next_node.next_node != list_tail:
p1 = p1.next_node
p1 = p1.next_node
p0 = p0.next_node
LinkedListNode.reverse_print(p0.next_node, list_tail)
LinkedListNode.reverse_print(list_head, p0)
if __name__ == "__main__":
list_head = LinkedListNode(4, LinkedListNode(5, LinkedListNode(12, LinkedListNode(1, LinkedListNode(3, None)))))
LinkedListNode.reverse_print(list_head, None)
這是O(n)時間和O(sqrt(n))空間算法。 在帖子的第二部分,它將擴展到線性時間和O(n ^(1 / t))空間算法,用於任意正整數t。
高級想法:將列表拆分為sqrt(n)許多(幾乎)相等大小的部分。 使用從最后到第一個的朴素線性時間線性空間方法以相反的順序依次打印部件。
要存儲部件的起始節點,我們需要一個大小為O(sqrt(n))的數組。 要恢復大小約為sqrt(n)的部分,朴素算法需要一個數組來存儲對部件節點的引用。 所以數組的大小為O(sqrt(n)。
一個使用大小為k=[sqrt(n)]+1 =O(sqrt(n))
兩個數組( lsa
和ssa
)(lsa ...大步長數組,ssa ...小步長數組)
階段1 :(如果鏈接列表的大小未知,則找出n,其長度):從頭到尾遍歷列表並計算列表的元素,這需要n個步驟
階段2:將單個鏈表的每個第k個節點存儲在陣列
lsa
。 這需要n個步驟。階段3:以相反的順序處理
lsa
列表。 以相反的順序打印每個部分這也需要n個步驟
因此算法的運行時間為3n = O(n),其速度約為2 * sqrt(n)= O(sqrt(n))。
這是一個Python 3.5實現:
import cProfile
import math
class LinkedListNode:
def __init__(self, value, next_node):
self.value = value
self._next_node = next_node
def next_node(self):
return(self._next_node)
def reverse_print(self):
# Phase 1
n=0
node=self
while node:
n+=1
node=node.next_node()
k=int(n**.5)+1
# Phase 2
i=0
node=self
lsa=[node]
while node:
i+=1
if i==k:
lsa.append(node)
i=0
last_node=node
node=node.next_node()
if i>0:
lsa.append(last_node)
# Phase 3
start_node=lsa.pop()
print(start_node.value)
while lsa:
last_printed_node=start_node
start_node=lsa.pop()
node=start_node
ssa=[]
while node!=last_printed_node:
ssa.append(node)
node=node.next_node()
ssa.reverse()
for node in ssa:
print(node.value)
@classmethod
def create_testlist(nodeclass, n):
''' creates a list of n elements with values 1,...,n'''
first_node=nodeclass(n,None)
for i in range(n-1,0,-1):
second_node=first_node
first_node=nodeclass(i,second_node)
return(first_node)
if __name__ == "__main__":
n=1000
cProfile.run('LinkedListNode.create_testlist(n).reverse_print()')
print('estimated number of calls of next_node',3*n)
它打印以下輸出(最后是分析器的輸出,顯示函數調用的數量):
>>>
RESTART: print_reversed_list3.py
1000
999
998
...
4
3
2
1
101996 function calls in 2.939 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 2.939 2.939 <string>:1(<module>)
2000 0.018 0.000 2.929 0.001 PyShell.py:1335(write)
1 0.003 0.003 2.938 2.938 print_reversed_list3.py:12(reverse_print)
1 0.000 0.000 0.001 0.001 print_reversed_list3.py:49(create_testlist)
1000 0.000 0.000 0.000 0.000 print_reversed_list3.py:5(__init__)
2999 0.000 0.000 0.000 0.000 print_reversed_list3.py:9(next_node)
...
estimated number of calls of next_node 3000
>>>
對next_node()的調用次數是預期的3000
不是使用朴素的O(m)空間算法以相反的順序打印長度為m的子列表,而是可以使用該O(sqrt(m))空間算法。 但我們必須在子列表的數量和子列表的長度之間找到適當的平衡:
階段2:將簡單鏈接列表拆分為長度為n ^(2/3)的n ^(1/3)子列表。 這些子列表的起始節點存儲在長度為n ^(1/3)的數組中
階段3:使用O(sqrt(m))空間算法以相反的順序打印長度m = n ^(2/3)的每個子列表。 因為m = n ^(2/3),我們需要m ^(1/2)= n ^(1/3)空間。
我們現在有一個需要4n時間的O(n ^(1/3))空間算法,因此仍然是O(n)
我們可以通過分裂成長度m = n ^(3/4)的n ^(1/4)子列表再次重復這一點,由O(m ^(1/3))= O(n ^(1 / 4))需要5n = O(n)時間的空間算法。
我們可以一次又一次地重復這一點,並得出以下聲明:
大小為n的不可變簡單鏈表可以使用t * n ^(1 / t)= O(n ^(1 / t))空間和(t + 1)n = O(n)時間以相反順序打印t是任意正整數
如果一個不修復t但是選擇t取決於n,使得n ^(1 / t))大約2,最小的有用數組大小,則這導致O(nlog(n))時間和O(log(n) ))由OP描述的空間算法。
如果選擇t = 1,則導致O(n)時間和O(n)空間天真算法。
這是算法的實現
import cProfile
import math
import time
class LinkedListNode:
'''
single linked list node
a node has a value and a successor node
'''
stat_counter=0
stat_curr_space=0
stat_max_space=0
stat_max_array_length=0
stat_algorithm=0
stat_array_length=0
stat_list_length=0
stat_start_time=0
do_print=True
def __init__(self, value, next_node):
self.value = value
self._next_node = next_node
def next_node(self):
self.stat_called_next_node()
return(self._next_node)
def print(self):
if type(self).do_print:
print(self.value)
def print_tail(self):
node=self
while node:
node.print()
node=node.next_node()
def tail_info(self):
list_length=0
node=self
while node:
list_length+=1
last_node=node
node=node.next_node()
return((last_node,list_length))
def retrieve_every_n_th_node(self,step_size,list_length):
''' for a list a of size list_length retrieve a pair there the first component
is an array with the nodes
[a[0],a[k],a[2*k],...,a[r*k],a[list_length-1]]]
and the second component is list_length-1-r*k
and
'''
node=self
arr=[]
s=step_size
index=0
while index<list_length:
if s==step_size:
arr.append(node)
s=1
else:
s+=1
last_node=node
node=node.next_node()
index+=1
if s!=1:
last_s=s-1
arr.append(last_node)
else:
last_s=step_size
return(arr,last_s)
def reverse_print(self,algorithm=0):
(last_node,list_length)=self.tail_info()
assert(type(algorithm)==int)
if algorithm==1:
array_length=list_length
elif algorithm==0:
array_length=2
elif algorithm>1:
array_length=math.ceil(list_length**(1/algorithm))
if array_length<2:
array_length=2
else:
assert(False)
assert(array_length>=2)
last_node.print()
self.stat_init(list_length=list_length,algorithm=algorithm,array_length=array_length)
self._reverse_print(list_length,array_length)
assert(LinkedListNode.stat_curr_space==0)
self.print_statistic()
def _reverse_print(self,list_length,array_length):
'''
this is the core procedure of the algorithm
if the list fits into the array
store it in te array an print the array in reverse order
else
split the list in 'array_length' sublists and store
the startnodes of the sublists in he array
_reverse_print array in reverse order
'''
if list_length==3 and array_length==2: # to avoid infinite loop
array_length=3
step_size=math.ceil(list_length/array_length)
if step_size>1: # list_length>array_length:
(supporting_nodes,last_step_size)=self.retrieve_every_n_th_node(step_size,list_length)
self.stat_created_array(supporting_nodes)
supporting_nodes.reverse()
supporting_nodes[1]._reverse_print(last_step_size+1,array_length)
for node in supporting_nodes[2:]:
node._reverse_print(step_size+1,array_length)
self.stat_removed_array(supporting_nodes)
else:
assert(step_size>0)
(adjacent_nodes,last_step_size)=self.retrieve_every_n_th_node(1,list_length)
self.stat_created_array(adjacent_nodes)
adjacent_nodes.reverse()
for node in adjacent_nodes[1:]:
node.print()
self.stat_removed_array(adjacent_nodes)
# statistics functions
def stat_init(self,list_length,algorithm,array_length):
'''
initializes the counters
and starts the stop watch
'''
type(self)._stat_init(list_length,algorithm,array_length)
@classmethod
def _stat_init(cls,list_length,algorithm,array_length):
cls.stat_curr_space=0
cls.stat_max_space=0
cls.stat_counter=0
cls.stat_max_array_length=0
cls.stat_array_length=array_length
cls.stat_algorithm=algorithm
cls.stat_list_length=list_length
cls.stat_start_time=time.time()
def print_title(self):
'''
prints the legend and the caption for the statistics values
'''
type(self).print_title()
@classmethod
def print_title(cls):
print(' {0:10s} {1:s}'.format('space','maximal number of array space for'))
print(' {0:10s} {1:s}'.format('', 'pointers to the list nodes, that'))
print(' {0:10s} {1:s}'.format('', 'is needed'))
print(' {0:10s} {1:s}'.format('time', 'number of times the method next_node,'))
print(' {0:10s} {1:s}'.format('', 'that retrievs the successor of a node,'))
print(' {0:10s} {1:s}'.format('', 'was called'))
print(' {0:10s} {1:s}'.format('alg', 'algorithm that was selected:'))
print(' {0:10s} {1:s}'.format('', '0: array size is 2'))
print(' {0:10s} {1:s}'.format('', '1: array size is n, naive algorithm'))
print(' {0:10s} {1:s}'.format('', 't>1: array size is n^(1/t)'))
print(' {0:10s} {1:s}'.format('arr', 'dimension of the arrays'))
print(' {0:10s} {1:s}'.format('sz', 'actual maximal dimension of the arrays'))
print(' {0:10s} {1:s}'.format('n', 'list length'))
print(' {0:10s} {1:s}'.format('log', 'the logarithm to base 2 of n'))
print(' {0:10s} {1:s}'.format('n log n', 'n times the logarithm to base 2 of n'))
print(' {0:10s} {1:s}'.format('seconds', 'the runtime of the program in seconds'))
print()
print('{0:>10s} {1:>10s} {2:>4s} {3:>10s} {4:>10s} {5:>10s} {6:>5s} {7:>10s} {8:>10s}'
.format('space','time','alg','arr','sz','n','log', 'n log n','seconds'))
@classmethod
def print_statistic(cls):
'''
stops the stop watch and prints the statistics for the gathered counters
'''
run_time=time.time()-cls.stat_start_time
print('{0:10d} {1:10d} {2:4d} {3:10d} {4:10d} {5:10d} {6:5d} {7:10d} {8:10.2f}'.format(
cls.stat_max_space,cls.stat_counter,cls.stat_algorithm,
cls.stat_array_length,cls.stat_max_array_length,cls.stat_list_length,
int(math.log2(cls.stat_list_length)),int(cls.stat_list_length*math.log2(cls.stat_list_length)),
run_time
))
def stat_called_next_node(self):
'''
counter: should be called
if the next node funtion is called
'''
type(self)._stat_called_next_node()
@classmethod
def _stat_called_next_node(cls):
cls.stat_counter+=1
def stat_created_array(self,array):
'''
counter: should be called
after an array was created and filled
'''
type(self)._stat_created_array(array)
@classmethod
def _stat_created_array(cls,array):
cls.stat_curr_space+=len(array)
if cls.stat_curr_space> cls.stat_max_space:
cls.stat_max_space=cls.stat_curr_space
if (len(array)>cls.stat_max_array_length):
cls.stat_max_array_length=len(array)
def stat_removed_array(self,array):
'''
counter: should be called
before an array can be removed
'''
type(self)._stat_removed_array(array)
@classmethod
def _stat_removed_array(cls,array):
cls.stat_curr_space-=len(array)
@classmethod
def create_testlist(nodeclass, n):
'''
creates a single linked list of
n elements with values 1,...,n
'''
first_node=nodeclass(n,None)
for i in range(n-1,0,-1):
second_node=first_node
first_node=nodeclass(i,second_node)
return(first_node)
if __name__ == "__main__":
#cProfile.run('LinkedListNode.create_testlist(n).reverse_print()')
n=100000
ll=LinkedListNode.create_testlist(n)
LinkedListNode.do_print=False
ll.print_title()
ll.reverse_print(1)
ll.reverse_print(2)
ll.reverse_print(3)
ll.reverse_print(4)
ll.reverse_print(5)
ll.reverse_print(6)
ll.reverse_print(7)
ll.reverse_print(0)
以下是一些結果
space maximal number of array space for
pointers to the list nodes, that
is needed
time number of times the method next_node,
that retrievs the successor of a node,
was called
alg algorithm that was selected:
0: array size is 2
1: array size is n, naive algorithm
t>1: array size is n^(1/t)
arr dimension of the arrays
sz actual maximal dimension of the arrays
n list length
log the logarithm to base 2 of n
n log n n times the logarithm to base 2 of n
seconds the runtime of the program in seconds
space time alg arr sz n log n log n seconds
100000 100000 1 100000 100000 100000 16 1660964 0.17
635 200316 2 317 318 100000 16 1660964 0.30
143 302254 3 47 48 100000 16 1660964 0.44
75 546625 4 18 19 100000 16 1660964 0.99
56 515989 5 11 12 100000 16 1660964 0.78
47 752976 6 7 8 100000 16 1660964 1.33
45 747059 7 6 7 100000 16 1660964 1.23
54 1847062 0 2 3 100000 16 1660964 3.02
和
space maximal number of array space for
pointers to the list nodes, that
is needed
time number of times the method next_node,
that retrievs the successor of a node,
was called
alg algorithm that was selected:
0: array size is 2
1: array size is n, naive algorithm
t>1: array size is n^(1/t)
arr dimension of the arrays
sz actual maximal dimension of the arrays
n list length
log the logarithm to base 2 of n
n log n n times the logarithm to base 2 of n
seconds the runtime of the program in seconds
space time alg arr sz n log n log n seconds
1000000 1000000 1 1000000 1000000 1000000 19 19931568 1.73
2001 3499499 2 1000 1001 1000000 19 19931568 7.30
302 4514700 3 100 101 1000000 19 19931568 8.58
131 4033821 4 32 33 1000000 19 19931568 5.69
84 6452300 5 16 17 1000000 19 19931568 11.04
65 7623105 6 10 11 1000000 19 19931568 13.26
59 7295952 7 8 9 1000000 19 19931568 11.07
63 21776637 0 2 3 1000000 19 19931568 34.39
就該問題的空間/時間要求而言,頻譜有兩端:
既然你不關心O(n)空間解決方案,讓我們看看另一個:
def reverse_print(LL):
length = 0
curr = LL
while curr:
length += 1
curr = curr.next
for i in range(length, 0, -1):
curr = LL
for _ in range(i):
curr = curr.next
print(curr.value)
當然,如果您選擇將其轉換為雙向鏈接列表,則可以在O(n)時間和0空間中執行此操作
渴望評論:
OP中算法的運行時間不是O(n)。 它是O(n log(n))。 作為運行時間,我們定義了我們對節點的下一個節點進行測試的次數。 這在方法reverse_print的主體中的3個位置處明確地完成。 實際上它是在5個地方完成的:while-clause中的2個和while looop中的3個,但如果一個臨時保存值,它可以減少到3個。 while循環重復約n / 2次。 因此,reverse_print方法明確地獲取下一個節點3/2 * 2次。 它在while循環之后的reverse_print的兩次調用中隱含地獲取它們。 要在這些調用中處理的列表長度是用於原始調用reverse_print的列表長度的一半,因此它是n / 2。 因此,我們對運行時間有以下近似值:
t(n) = 1.5n+2t(n/2)
這種復發的解決方案是
t(n) = 1.5n log(n) + n
如果將解決方案插入到reccurence中,則可以驗證這一點。
您還可以運行問題計算獲取節點的頻率。 為此我向你的程序添加了一個next_node()方法。 我使用cProfiler來計算函數調用。 我還添加了一個類方法來創建測試列表。 最后以這個程序結束
import cProfile
import math
class LinkedListNode:
def __init__(self, value, next_node):
self.value = value
self._next_node = next_node
def next_node(self):
''' fetch the next node'''
return(self._next_node)
def reverse_print(self, list_tail):
list_head=self
if not self:
return
if not self.next_node():
print (self.value)
return
if self == list_tail:
print (self.value)
return
p0 = self
p1 = self
#assert(p1.next_node != list_tail)
p1_next=p1.next_node()
p1_next_next=p1_next.next_node()
while p1_next != list_tail and p1_next_next != list_tail:
p1 = p1_next_next
p0 = p0.next_node()
p1_next=p1.next_node()
if p1_next != list_tail:
p1_next_next=p1_next.next_node()
p0.next_node().reverse_print(list_tail)
self.reverse_print(p0)
@classmethod
def create_testlist(nodeclass, n):
''' creates a list of n elements with values 1,...,n'''
first_node=nodeclass(n,None)
for i in range(n-1,0,-1):
second_node=first_node
first_node=nodeclass(i,second_node)
return(first_node)
if __name__ == "__main__":
n=1000
cProfile.run('LinkedListNode.create_testlist(n).reverse_print(None)')
print('estimated number of calls of next_node',1.5*n*math.log(n,2)+n)
我得到了以下輸出(最后是探查器的輸出,顯示了函數調用的數量):
>>>
RESTART: print_reversed_list2.py
1000
999
998
...
2
1
116221 function calls (114223 primitive calls) in 2.539 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
1 0.000 0.000 2.539 2.539 <string>:1(<module>)
2000 0.015 0.000 2.524 0.001 PyShell.py:1335(write)
1999/1 0.008 0.000 2.538 2.538 print_reversed_list2.py:12(reverse_print)
1 0.000 0.000 0.001 0.001 print_reversed_list2.py:36(create_testlist)
1000 0.000 0.000 0.000 0.000 print_reversed_list2.py:5(__init__)
16410 0.002 0.000 0.002 0.000 print_reversed_list2.py:9(next_node)
...
estimated number of calls of next_node 15948.67642699313
因此,通過公式估計的next_node()調用的數量約為15949.send_node()調用的實際數量是16410.后一個數字包括行p0.next_node().reverse_print(list_tail)
的next_node()的2000次調用.reverse_print p0.next_node().reverse_print(list_tail)
我沒有考慮我的公式。
因此, 1.5*n*log(n)+n
似乎是對程序運行時間的合理估計。
免責聲明:我錯過了在本次討論的背景下無法修改列表。
想法:我們按正向順序迭代列表,在我們處於此狀態時將其反轉。 當我們到達終點時,我們向后迭代,打印元素並再次反轉列表。
核心觀察是你可以就地反轉一個列表:你需要的只是記住你最后一個元素。
未經測試,丑陋的偽代碼:
def printReverse(list) {
prev = nil
cur = list.head
if cur == nil {
return
}
while cur != nil {
next = cur.next
// [prev] cur -> next
cur.next = prev
// [prev] <- cur next
prev = cur
// [____] <- prev next
cur = next
// [____] <- prev cur
}
// Now cur is nil and prev the last element!
cur = prev
prev = nil
while cur != nil {
print cur
// Rewire just as above:
next = cur.next
cur.next = prev
prev = cur
cur = next
}
}
顯然,這在時間O(n)中運行並且僅占用O(1)(附加)空間(三個本地指針/引用變量)。
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