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04698538d8
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>
78 lines
1.8 KiB
Python
78 lines
1.8 KiB
Python
def is_palindrome(head):
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if not head:
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return True
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# split the list to two parts
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fast, slow = head.next, head
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while fast and fast.next:
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fast = fast.next.next
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slow = slow.next
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second = slow.next
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slow.next = None # Don't forget here! But forget still works!
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# reverse the second part
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node = None
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while second:
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nxt = second.next
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second.next = node
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node = second
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second = nxt
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# compare two parts
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# second part has the same or one less node
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while node:
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if node.val != head.val:
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return False
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node = node.next
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head = head.next
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return True
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def is_palindrome_stack(head):
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if not head or not head.next:
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return True
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# 1. Get the midpoint (slow)
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slow = fast = cur = head
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while fast and fast.next:
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fast, slow = fast.next.next, slow.next
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# 2. Push the second half into the stack
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stack = [slow.val]
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while slow.next:
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slow = slow.next
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stack.append(slow.val)
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# 3. Comparison
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while stack:
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if stack.pop() != cur.val:
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return False
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cur = cur.next
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return True
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def is_palindrome_dict(head):
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if not head or not head.next:
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return True
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d = {}
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pos = 0
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while head:
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if head.val in d:
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d[head.val].append(pos)
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else:
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d[head.val] = [pos]
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head = head.next
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pos += 1
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checksum = pos - 1
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middle = 0
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for v in d.values():
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if len(v) % 2 != 0:
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middle += 1
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else:
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step = 0
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for i in range(0, len(v)):
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if v[i] + v[len(v) - 1 - step] != checksum:
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return False
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step += 1
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if middle > 1:
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return False
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return True
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