问题
here is my code:
def merge_lists(head1, head2):
if head1 is None and head2 is None:
return None
if head1 is None:
return head2
if head2 is None:
return head1
if head1.value < head2.value:
temp = head1
else:
temp = head2
while head1 != None and head2 != None:
if head1.value < head2.value:
temp.next = head1
head1 = head1.next
else:
temp.next = head2
head2 = head2.next
if head1 is None:
temp.next = head2
else:
temp.next = head1
return temp
pass
the problem here is stucked in the infinite loop.can any one tell me what the problem is
the examples are:
assert [] == merge_lists([],[])
assert [1,2,3] == merge_lists([1,2,3], [])
assert [1,2,3] == merge_lists([], [1,2,3])
assert [1,1,2,2,3,3,4,5] == merge_lists([1,2,3], [1,2,3,4,5])
回答1:
The problem with the current code is that it causes a side-effect of the temp node's next before it navigates to the next node from the current node. This is problematic when the current temp node is the current node.
That is, imagine this case:
temp = N
temp.next = N # which means N.next = N
N = N.next # but from above N = (N.next = N) -> N = N
There is a corrected version, with some other updates:
def merge_lists(head1, head2):
if head1 is None:
return head2
if head2 is None:
return head1
# create dummy node to avoid additional checks in loop
s = t = node()
while not (head1 is None or head2 is None):
if head1.value < head2.value:
# remember current low-node
c = head1
# follow ->next
head1 = head1.next
else:
# remember current low-node
c = head2
# follow ->next
head2 = head2.next
# only mutate the node AFTER we have followed ->next
t.next = c
# and make sure we also advance the temp
t = t.next
t.next = head1 or head2
# return tail of dummy node
return s.next
回答2:
Recursive algorithm for merging two sorted linked lists
def merge_lists(h1, h2):
if h1 is None:
return h2
if h2 is None:
return h1
if (h1.value < h2.value):
h1.next = merge_lists(h1.next, h2)
return h1
else:
h2.next = merge_lists(h2.next, h1)
return h2
来源:https://stackoverflow.com/questions/22507197/merging-two-sorted-linked-lists-into-one-linked-list-in-python