def reverseList2(self, head: ListNode) -> ListNode:
if not head or head.next == None:
return head
res = self.reverseList2(head.next)
head.next.next = head
head.next = None
return res
def removeNthFromEnd3(self, head: ListNode, n: int) -> ListNode:
if not head:
self.count = 0
return head
head.next = self.removeNthFromEnd3(head.next, n)
self.count += 1
return head.next if self.count == n else head
def removeNthFromEnd2(self, head: ListNode, n: int) -> ListNode:
# dummy,添加头节点
dummy = ListNode(0)
dummy.next = head
# 快指针先走 n 步
slow, fast = dummy, dummy
for _ in range(n):
fast = fast.next
# 快慢指针同时走，直到 fast 指针达到链表尾部节点，此时 slow 到达倒数第 n 个节点的前一个节点
while fast and fast.next:
slow, fast = slow.next, fast.next
# 删除节点并重新链接
slow.next = slow.next.next
return dummy.next
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
# dummy,添加头节点
dummy = ListNode(0)
dummy.next = head
# 获取链表长度
cur, lenth = head, 0
while cur:
lenth += 1
cur = cur.next
# 找到倒数第 n 个节点前的那个节点
cur = dummy
# 以示例为例，注意长度是 5 而不是 4
for _ in range(lenth - n):
cur = cur.next
# 删除节点并重新链接
cur.next = cur.next.next
return dummy.next
def getIntersectionNode(self, headA: ListNode,
headB: ListNode) -> ListNode:
""""""
nodeA, nodeB = headA, headB
while nodeA != nodeB:
nodeA = nodeA.next if nodeA else headB
nodeB = nodeB.next if nodeB else headA
return nodeA
if __name__ == '__main__':
s = Solution()
# [4, 1, 8, 4, 5]
# [5, 6, 1, 8, 4, 5]
nodei = ListNode(8)
nodei.next = ListNode(4)
nodei.next.next = ListNode(5)
headA = ListNode(4)
headA.next = ListNode(1)
headA.next.next = nodei
headB = ListNode(5)
headB.next = ListNode(6)
headB.next.next = ListNode(1)
headB.next.next.next = nodei
# [2,6,4]
# [1, 5]
# headA = ListNode(2)
# headA.next = ListNode(6)
sumNum = int(l1Num[::-1]) + int(l2Num[::-1])
r1 = str(sumNum)[::-1]
r2 = list(map(lambda x: ListNode(x), r1))
for i in range(len(r1) - 1):
r2[i].next = r2[i + 1]
node = r2[0]
return node
# Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
# Output: 7 -> 0 -> 8
n2 = ListNode(2)
n4 = ListNode(4)
n3 = ListNode(3)
n2.next = n4
n4.next = n3
l5 = ListNode(5)
l6 = ListNode(6)
l4 = ListNode(4)
l5.next = l6
l6.next = l4
r = addTwoNums(n2, l5)
while r is not None:
print(r.val)
r = r.next