def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
"""worst"""
if l1 is None:
return l2
if l2 is None:
return l1
l1_node = l1
l2_node = l2
vals = []
while l1_node is not None:
vals.append(l1_node.val)
l1_node = l1_node.next
while l2_node is not None:
vals.append(l2_node.val)
l2_node = l2_node.next
vals = sorted(vals)
return createList(vals)
for _ in range(mid_pos - 1):
node = node.next
return node
def middleNode(self, head: ListNode) -> ListNode:
"""
Time O(n)
Space O(1)
"""
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow
if __name__ == "__main__":
s = Solution()
data = [
([1, 2, 3, 4, 5], 3),
([1, 2, 3, 4, 5, 6], 4),
([1], 1),
([1, 2], 2),
([1, 2, 3], 2),
]
for it, gt in data:
assert (
s.middleNode2(createList(it)).val == gt
), f"{it} {s.middleNode2(createList(it)).val} {gt}"
class Solution:
def oddEvenList(self, head: ListNode) -> ListNode:
if head is None:
return None
odd = head
even = head.next
evenHead = even
while even is not None and even.next is not None:
odd.next = even.next
odd = odd.next
even.next = odd.next
even = even.next
odd.next = evenHead
return head
if __name__ == "__main__":
s = Solution()
head = createList([1, 2, 3, 4, 5])
out = s.oddEvenList(head)
assert compareList(out, [1, 3, 5, 2, 4])
head = createList([2, 1, 3, 5, 6, 4, 7])
out = s.oddEvenList(head)
assert compareList(out, [2, 3, 6, 7, 1, 5, 4])
Time: O(n)
Space: O(1)
"""
curr = head
while (curr is not None) and (curr.next is not None):
if curr.val != curr.next.val:
curr = curr.next
else:
# curr 不动，只改 curr 的 next，相当于跳过了当中一个
curr.next = curr.next.next
return head
if __name__ == '__main__':
s = Solution()
l1 = createList([1])
assert compareList(s.deleteDuplicates(l1), [1]) == True
l1 = createList([1, 1, 2])
assert compareList(s.deleteDuplicates(l1), [1, 2]) == True
l1 = createList([1, 1, 2, 3, 3])
assert compareList(s.deleteDuplicates(l1), [1, 2, 3]) == True
l1 = createList([1, 2, 3, 4, 5])
assert compareList(s.deleteDuplicates(l1), [1, 2, 3, 4, 5]) == True
l1 = createList([1, 1, 1, 1, 1])
assert compareList(s.deleteDuplicates(l1), [1]) == True
l1 = createList([1])
if n == 1:
self.successor = head.next
return head
last = self.reverseN(head.next, n - 1)
head.next.next = head
head.next = self.successor
return last
def reverseBetween(self, head: Optional[ListNode], left: int,
right: int) -> Optional[ListNode]:
"""
OMG! https://leetcode.com/problems/reverse-linked-list-ii/solution/242639
Time: O(n)
Space: O(1)
"""
if left == 1:
reversed_N = self.reverseN(head, right)
return reversed_N
head.next = self.reverseBetween(head.next, left - 1, right - 1)
return head
if __name__ == "__main__":
s = Solution()
data = [
([1, 2, 3, 4, 5], 2, 4, [1, 4, 3, 2, 5]),
]
for it, l, r, gt in data:
assert compareList(s.reverseBetween(createList(it), l, r), gt) is True
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
"""
Time: O(n)
Space: O(1)
"""
last_node = None
while head is not None:
next_head = head.next
head.next = last_node
last_node = head
head = next_head
return last_node
if __name__ == "__main__":
s = Solution()
data = [
[1],
[1, 2],
[1, 2, 3],
[1, 2, 3, 4, 5],
]
for it in data:
it_reversed = it.copy()
it_reversed.reverse()
assert compareList(s.reverseList(createList(it)), it_reversed) is True
slow.next = slow_head
slow_head = slow
slow = next_slow_head
while slow_head:
if slow_head.val != head.val:
return False
slow_head = slow_head.next
head = head.next
return True
if __name__ == "__main__":
s = Solution()
head = createList([1, 2, 2, 1])
assert s.isPalindrome(head) == True
head = createList([1, 2, 2])
assert s.isPalindrome(head) == False
head = createList([2, 2])
assert s.isPalindrome(head) == True
assert s.isPalindrome(None) == True
head = createList([1])
assert s.isPalindrome(head) == True
head = createList([1, 2, 2, 1])
assert s.isPalindrome2(head) == True
def compare(s, l1, l2, result):
l1 = createList(l1)
l2 = createList(l2)
out = s.addTwoNumbers(l1, l2)
print(out)
assert compareList(out, result) == True
# @Time : 2020/10/23 11:13
# @Author : LiuBin
# @File : 234.py
# @Description :
# @Software: PyCharm
"""回文链表
关键字: 栈
思路:
1、利用栈记录链表的倒序序列
2、依次比较栈和链表的值是否相等即可
"""
from utils import ListNode, createList
class Solution:
def isPalindrome(self, head: ListNode) -> bool:
stack = []
root = head
while head:
stack.append(head.val)
head = head.next
while root:
if stack.pop() != root.val:
return False
root = root.next
return True
print(Solution().isPalindrome(createList([1, 2, 1])))
# Space: O(1)
dummy = cur = ListNode(0)
while l1 and l2:
if l1.val < l2.val:
cur.next = l1
l1 = l1.next
else:
cur.next = l2
l2 = l2.next
cur = cur.next
cur.next = l1 or l2
return dummy.next
if __name__ == '__main__':
s = Solution()
l1 = createList([1, 2, 4])
l2 = createList([1, 3, 4])
assert compareList(l2, [1, 3, 4]) == True
assert compareList(l2, [1, 2, 4]) == False
assert compareList(s.mergeTwoLists1(l1, l2), [1, 1, 2, 3, 4, 4]) == True
l1 = createList([1, 2, 4])
l2 = createList([1])
assert compareList(s.mergeTwoLists1(l1, l2), [1, 1, 2, 4]) == True
l1 = createList([1, 2, 4])
l2 = createList([5])
assert compareList(s.mergeTwoLists1(l1, l2), [1, 2, 4, 5]) == True
def test(l1, l2, result):
l1 = createList(l1)
l2 = createList(l2)
assert compareList(s.addTwoNumbers(l1, l2), result) == True
break
else:
head = head.next
prev = None
curr = head
while curr is not None:
if curr.val == val:
# prev 为 head 的引用，这一步相当于在原始 llinked list 上操作
prev.next = curr.next
curr = curr.next
else:
prev = curr
curr = curr.next
return head
if __name__ == "__main__":
s = Solution()
head = createList([1, 2, 6, 3, 4, 5, 6])
out = s.removeElements(head, 6)
assert compareList(out, [1, 2, 3, 4, 5])
head = createList([6, 2, 6, 3, 4, 5, 6])
out = s.removeElements(head, 6)
assert compareList(out, [2, 3, 4, 5])
head = createList([6, 6, 6])
out = s.removeElements(head, 6)
assert compareList(out, [])
# if they meet, pa or pb would be the node we are looking for,
# if they didn't meet, they will hit the end at the same iteration, pa == pb == None, return either one of them is the same,None
if headA is None or headB is None:
return None
pa = headA # 2 pointers
pb = headB
while pa is not pb:
# if either pointer hits the end, switch head and continue the second traversal,
# if not hit the end, just move on to next
pa = headB if pa is None else pa.next
pb = headA if pb is None else pb.next
return pa # only 2 ways to get out of the loop, they meet or the both hit the end=None
if __name__ == '__main__':
s = Solution()
l1 = createList([1, 2, 3, 4, 5])
l2 = createList([8, 3, 4, 5])
assert compareList(s.getIntersectionNode(l1, l2), [3, 4, 5]) == True
l1 = createList([1, 3, 5, 7, 9, 11])
l2 = createList([2, 4, 9, 11])
assert compareList(s.getIntersectionNode(l1, l2), [9, 11]) == True
l1 = createList([1, 2, 3])
l2 = createList([1, 2])
assert compareList(s.getIntersectionNode(l1, l2), None) == True