while fast and fast.next:
slow = slow.next
fast = fast.next.next
# 反转链表
p = slow.next
slow.next = None
if p is None: return
pre = None
while p:
aft = p.next
p.next = pre
pre = p
p = aft
# 合并链表
a, b = head, pre
while True:
an = a.next
bn = b.next if b else None
a.next = b
if b:
b.next = an
a = an
b = bn
if not b:
break
node = ListNode.fromList([1])
Solution().reorderList(node)
print(node)
for _ in range(sub_len - 1):
if cur and cur.next:
cur = cur.next
else:
break
h2 = cur.next
cur.next = None # h1 结扎
cur = h2 # 恢复 cur
for _ in range(sub_len - 1):
if cur and cur.next:
cur = cur.next
else:
break
suc = None
if cur: # 构造新的【cur】
suc = cur.next
cur.next = None # h2 结扎
cur = suc #
merged = merge(h1, h2)
pre.next = merged
# 构造新的【pre】 (merged的最后一个结点)
while pre.next:
pre = pre.next
sub_len *= 2
return dummy.next
print(Solution().sortList(ListNode.fromList([4, 2, 1, 3])))
#
# print(merge(ListNode.fromList([1, 3, 5]), ListNode.fromList([2, 4, 6, 7])))
from structure import ListNode
class Solution:
def isPalindrome(self, head: ListNode) -> bool:
# hash = hash * seed + val
# seed: prime number
# val: node value
# hash1 = a0 * seed^(n-1) + a1 * seed^(n-2)
hash1 = hash2 = 0
h = 1
# seed=int(1e9+7)
seed = 3
p = head
while p is not None:
hash1 = hash1 * seed + p.val
hash2 = hash2 + h * p.val
h *= seed
p = p.next
return hash1 == hash2
if __name__ == '__main__':
node = ListNode.fromList([1, 2, 2, 1])
print(Solution().isPalindrome(node))
#!/usr/bin/env python # -*- coding: utf-8 -*- # @Author : qichun tang # @Contact : [email protected] from structure import ListNode class Solution: def oddEvenList(self, head: ListNode) -> ListNode: if not head: return head oddHead, evenHead = head, head.next odd, even = oddHead, evenHead while even and even.next: odd.next = even.next odd = odd.next # 别忘了自己要移动 even.next = odd.next even = even.next odd.next = evenHead return head if __name__ == '__main__': nodes = ListNode.fromList([1, 2, 3, 4, 5]) solved = Solution().oddEvenList(nodes) print(solved)
def insertionSortList(self, head: ListNode) -> ListNode:
if not head: return head
dummy = ListNode(-inf) # D
dummy.next = head # D 4 2 1 3 ∅
tail = dummy.next # 4 ∅ // 如果新结点比【有序区】的【尾结点】更大，直接成为新的尾结点
p = head.next # 2 1 3 ∅ // 头结点的第一个元素初始为【有序区】，从头结点的【下一个元素】开始遍历
tail.next = None # 封闭有序区
while p:
aft = p.next # 缓存下一个结点
if p.val >= tail.val: # 注意符号
tail.next = p
tail = p
else:
dp = dummy
while dp.next:
# 首次出现比p大的元素。插入到那个元素前面
if dp.next.val > p.val:
p.next = dp.next
dp.next = p
break
dp=dp.next
p = aft
tail.next = None
return dummy.next
head = ListNode.fromList([4, 2, 1, 3])
ans = Solution().insertionSortList(head)
print(ans)
# 递归终点
if not head or not head.next:
return head
# 中点寻找
slow = head
fast = head.next # 否则爆栈
while fast and fast.next:
slow = slow.next
fast = fast.next.next
mid = slow.next
slow.next = None
# 处理左链表与右链表
left = self.sortList(head)
right = self.sortList(mid)
# 合并两个排序连败哦
dummy = dp = ListNode(0)
while left and right:
if left.val < right.val:
dp.next = left
left = left.next
else:
dp.next = right
right = right.next
dp = dp.next
dp.next = left if left else right
return dummy.next
ans = Solution().sortList(ListNode.fromList([1, 5, 3, 4, 0]))
print(ans)
from structure import ListNode
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
if head is None or head.next is None:
return head
node = self.reverseList(head.next)
head.next.next = head
head.next = None
return node
print(Solution().reverseList(ListNode.fromList([1, 2, 3, 4, 5])))
from structure import ListNode
class Solution:
def swapPairs(self, head: ListNode) -> ListNode:
if not head: return None
dummy = ListNode()
dp = dummy
p = head
while p:
np = p.next
if np:
nnp = np.next
dp.next = np
dp = dp.next
dp.next = p
dp = dp.next
p = nnp
else:
dp.next = p
dp = dp.next
break
if dp:
dp.next = None
return dummy.next
print(Solution().swapPairs(ListNode.fromList([1])))
from structure import ListNode
node = ListNode.fromList([1, 2])
def partition(head: ListNode):
if not head or not head.next:
return head
p1, p2 = head, head.next
head1, head2 = p1, p2
# p2 后面至少有1个结点
while p2 and p2.next:
p1.next = p2.next
p1 = p1.next
p2.next = p1.next
p2 = p2.next
p1.next = None
return head1, head2
print(partition(node))
from structure import ListNode
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
p = head
k = 0
p2 = None
while p is not None:
p = p.next
k += 1
if (k - 1) == n:
p2 = head
if (k - 1) > n:
p2 = p2.next
if p2 is None and k == n:
head = head.next
elif p2.next is not None:
p2.next = p2.next.next
return head
if __name__ == '__main__':
node = Solution().removeNthFromEnd(ListNode.fromList([1, 2]), 2)
print(node)
class Solution:
def reorderList(self, head: ListNode) -> None:
if head is None:
return None
p = head
vec = []
while p is not None:
vec.append(p)
p = p.next
i = 0
j = len(vec) - 1
while i < j:
vec[i].next = vec[j]
i += 1
if i == j:
break
vec[j].next = vec[i]
j -= 1
vec[i].next = None # 容易想错
node = ListNode.fromList([1, 2, 3, 4])
Solution().reorderList(node)
print(node)
node = ListNode.fromList([1, 2, 3, 4, 5])
Solution().reorderList(node)
print(node)
right: int) -> ListNode:
# 以case 1, 2, 3, 4, 5 为例，left = 2, right = 4
p = head
L = right - left + 1 # 翻转的部分是 2 3 4， 长度为3
dummy = ListNode(0)
dp = dummy
for _ in range(left - 1): # 把 翻转部分 2 之前的添加到dummy中（dummy用来返回）
dp.next = p
dp = dp.next
p = p.next
# 迭代完之后 p=2
aft = p.next
pre = None
l2_tail = p # 第二段（l2） 的开头，但翻转后会变成末尾
for _ in range(L):
aft = p.next
p.next = pre
pre = p
p = aft
dp.next = pre # 接上翻转的列表， 即 1 4 3 2
l2_tail.next = aft # l2 段 接上剩下没迭代的
return dummy.next
case1 = ListNode.fromList([3, 5])
ans = Solution().reverseBetween(case1, 1, 2)
print(ans)
# ans = Solution().reverseBetween(ListNode.fromList(list(range(1, 6))), 2, 4)
# print(ans)
def reverse(a, b):
pre = None
p = a
while p != b:
aft = p.next
p.next = pre
pre = p
p = aft
return pre
class Solution:
def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
if head is None:
return None
a = head
b = head
for _ in range(k):
if b is None:
return head
b = b.next
n_head = reverse(a, b)
a.next = self.reverseKGroup(b, k)
return n_head
ans = Solution().reverseKGroup(ListNode.fromList([1, 2, 3, 4, 5]), 2)
print(ans)
if a.val < b.val:
dp.next = a
a = a.next
else:
dp.next = b
b = b.next
dp = dp.next
dp.next = a if a else b
return dummy.next
def partition(lists):
n = len(lists)
if n == 0:
return None
elif n == 1:
return lists[0]
else:
mid = len(lists) // 2
return merge(partition(lists[:mid]), partition(lists[mid:]))
class Solution:
def mergeKLists(self, lists: List[ListNode]) -> ListNode:
return partition(lists)
a = ListNode.fromList([1, 3, 5])
b = ListNode.fromList([2, 4])
print(merge(a, b))