Python/data_structures/linked_list/is_palindrome.py
CenTdemeern1 04698538d8
Misc fixes across multiple algorithms (#6912)
Source: Snyk code quality
Add scikit-fuzzy to requirements

Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
Co-authored-by: Dhruv Manilawala <dhruvmanila@gmail.com>
2022-10-16 10:55:38 +05:30

78 lines
1.8 KiB
Python

def is_palindrome(head):
if not head:
return True
# split the list to two parts
fast, slow = head.next, head
while fast and fast.next:
fast = fast.next.next
slow = slow.next
second = slow.next
slow.next = None # Don't forget here! But forget still works!
# reverse the second part
node = None
while second:
nxt = second.next
second.next = node
node = second
second = nxt
# compare two parts
# second part has the same or one less node
while node:
if node.val != head.val:
return False
node = node.next
head = head.next
return True
def is_palindrome_stack(head):
if not head or not head.next:
return True
# 1. Get the midpoint (slow)
slow = fast = cur = head
while fast and fast.next:
fast, slow = fast.next.next, slow.next
# 2. Push the second half into the stack
stack = [slow.val]
while slow.next:
slow = slow.next
stack.append(slow.val)
# 3. Comparison
while stack:
if stack.pop() != cur.val:
return False
cur = cur.next
return True
def is_palindrome_dict(head):
if not head or not head.next:
return True
d = {}
pos = 0
while head:
if head.val in d:
d[head.val].append(pos)
else:
d[head.val] = [pos]
head = head.next
pos += 1
checksum = pos - 1
middle = 0
for v in d.values():
if len(v) % 2 != 0:
middle += 1
else:
step = 0
for i in range(0, len(v)):
if v[i] + v[len(v) - 1 - step] != checksum:
return False
step += 1
if middle > 1:
return False
return True