newCurrent.next = current
newCurrent = newCurrent.next
newCurrent.next = None
current = next
return newHead.next
def printList(l: ListNode):
while l:
print(l.val, flush=True, sep=' ', end=' ')
l = l.next
print()
l1 = ListNode(1)
l1.next = ListNode(2)
l1.next.next = ListNode(3)
l1.next.next.next = ListNode(3)
l1.next.next.next.next = ListNode(4)
l1.next.next.next.next.next = ListNode(4)
l1.next.next.next.next.next.next = ListNode(5)
printList(Solution().deleteDuplicates(l1)) # 1 2 5
l2 = ListNode(1)
l2.next = ListNode(1)
l2.next.next = ListNode(1)
l2.next.next.next = ListNode(2)
l2.next.next.next.next = ListNode(3)
printList(Solution().deleteDuplicates(l2)) # 2 3
Memory Usage: 15.3 MB, less than 62.20% of Python3 online submissions for Reverse Linked List.
"""
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
curr = head
prev: ListNode = None
while curr:
next = curr.next
curr.next = prev
prev = curr
curr = next
return prev
def printList(l: ListNode):
while l:
print(l.val, flush=True, sep=' ', end=' ')
l = l.next
print()
l = ListNode(1)
l.next = ListNode(2)
l.next.next = ListNode(3)
l.next.next.next = ListNode(4)
l.next.next.next.next = ListNode(5)
printList(Solution().reverseList(l)) # 5 4 3 2 1
headA = advance(headA, cnt)
if len2 > len1:
cnt = len2 - len1
headB = advance(headB, cnt)
while headA and headB:
if headA == headB:
return headA
headA = headA.next
headB = headB.next
return None
l1a = ListNode(4)
l1a.next = ListNode(1)
l1a.next.next = ListNode(8)
l1a.next.next.next = ListNode(4)
l1a.next.next.next.next = ListNode(5)
l1b = ListNode(5)
l1b.next = ListNode(6)
l1b.next.next = ListNode(1)
l1b.next.next.next = l1a.next.next
print(Solution().getIntersectionNode(l1a, l1b).val) # 8
l2a = ListNode(1)
l2a.next = ListNode(9)
l2a.next.next = ListNode(1)
l2a.next.next.next = ListNode(2)
l2a.next.next.next.next = ListNode(4)
current.next = tail
next.next = current
if prev:
prev.next = next
prev = current
current = tail
return newHead
def printList(l: ListNode):
while l:
print(l.val, flush=True, sep=' ', end=' ')
l = l.next
print()
list1 = ListNode(1)
list1.next = ListNode(2)
list1.next.next = ListNode(3)
list1.next.next.next = ListNode(4)
list1.next.next.next.next = ListNode(5)
printList(Solution().swapPairs(list1))
list2 = ListNode(1)
list2.next = ListNode(2)
list2.next.next = ListNode(3)
list2.next.next.next = ListNode(4)
printList(Solution().swapPairs(list2))
curr: ListNode = head
while curr:
next = curr.next
curr.next = prev
prev = curr
curr = next
return prev
def isPalindrome(self, head: ListNode) -> bool:
if head is None:
return True
middle = self.getMiddle(head)
middle = self.reverse(middle)
while middle:
if middle.val != head.val:
return False
middle = middle.next
head = head.next
return True
l1 = ListNode(1)
l1.next = ListNode(2)
print(Solution().isPalindrome(l1)) # False
l2 = ListNode(1)
l2.next = ListNode(2)
l2.next.next = ListNode(2)
l2.next.next.next = ListNode(1)
print(Solution().isPalindrome(l2)) # True
prev = curr
curr = curr.next
if prev2:
prev2.next = None
def printList(l: ListNode):
while l:
print(l.val, flush=True, sep=' ', end=' ')
l = l.next
print()
l1 = ListNode(4)
n5 = l1.next = ListNode(5)
l1.next.next = ListNode(1)
l1.next.next.next = ListNode(9)
Solution().deleteNode(n5)
printList(l1) # 4 1 9
l2 = ListNode(4)
l2.next = ListNode(5)
n1 = l2.next.next = ListNode(1)
l2.next.next.next = ListNode(9)
Solution().deleteNode(n1)
printList(l2) # 4 5 9