from python3.aatool.LinkT import ListNode, createSLink
# @lc code=start
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
root, pre, x = None, None, 0
while l1 or l2 or x != 0:
v1 = l1.val if l1 else 0
v2 = l2.val if l2 else 0
x, y = divmod(v1 + v2 + x, 10)
node = ListNode(y, None)
if pre:
pre.next = node
else:
root = node
pre = node
l1 = l1.next if l1 else l1
l2 = l2.next if l2 else l2
return root
# @lc code=end
Solution().addTwoNumbers(createSLink([9, 9, 9, 9, 9, 9, 9]),
createSLink([9, 9, 9, 9]))
tailNode = self.recursion(head, head.next)
# head.next 保存新的头节点
preNode.next = head.next
# 单前节点移到尾部
tailNode.next, head.next = head, tailNode.next
return tailNode.next
else:
return head
# 迭代
def iteration(self, head:ListNode) -> ListNode:
# 空链，单节点链直接返回
if not head or not head.next: return head
# 从第二个节点开始遍历
headNode, tailNode, curNode = head, head, head.next
while curNode:
# 当前节点后一个节点移到最新尾部，最新表头移到当前节点后
tailNode.next, curNode.next = curNode.next, headNode
# 更新最新节点&单前节点
headNode, curNode = curNode, tailNode.next
return headNode
def reverseList(self, head: ListNode) -> ListNode:
return self.iteration(head)
# dummyNode = ListNode(val=0, next=head)
# self.recursion(dummyNode, head)
# return dummyNode.next
# @lc code=end
Solution().reverseList(createSLink([1,2,3,4,5]))
# 反转链表的k个节点并返回反转后最后一个节点
def reverse(self, preNode: ListNode, head: ListNode, k: int) -> ListNode:
temp, stack = head, []
for i in range(k):
if temp:
stack.append(temp)
temp = temp.next
else:
break
if len(stack) < k:
return None
for i in range(k-1, -1, -1):
preNode.next = stack[i]
preNode = preNode.next
if i == k-1:
lstNode = stack[i].next
if i == 0:
stack[i].next = lstNode
return preNode
def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
dummyNode = ListNode(val=0, next=head)
preNode = dummyNode
while preNode:
preNode = self.reverse(preNode, head, k)
head = preNode.next if preNode else None
return dummyNode.next
# @lc code=end
Solution().reverseKGroup(createSLink([1,2,3,4,5]), 1)
# curMap, preMap, preNode, tmp = [], [], None, head
# while tmp:
# curMap.append(tmp)
# preMap.append(preNode)
# preNode = tmp
# tmp = tmp.next
# dIndex = len(curMap) - n
# curNode = curMap[dIndex]
# preNode = preMap[dIndex]
# if preNode:
# preNode.next = curNode.next
# else:
# head = head.next
# return head
def removeNthFromEnd(self, head, n):
def index(node):
if not node:
return 0
i = index(node.next) + 1
if i > n:
node.next.val = node.val
return i
index(head)
return head.next
# @lc code=end
Solution().removeNthFromEnd(createSLink([1, 2, 3, 4, 5]), 2)
#
# @lc app=leetcode.cn id=203 lang=python3
#
# [203] 移除链表元素
#
from python3.aatool.LinkT import ListNode, createSLink
# @lc code=start
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeElements(self, head: ListNode, val: int) -> ListNode:
dummyNode = ListNode(val=0, next=head)
curNode = dummyNode
while curNode.next:
if curNode.next.val == val:
curNode.next = curNode.next.next
else:
curNode = curNode.next
return dummyNode.next
# @lc code=end
Solution().removeElements(createSLink([7, 7, 7, 7]), 7)
while curNode:
tailNode.next, curNode.next = curNode.next, headNode
headNode, curNode = curNode, tailNode.next
return headNode
def isPalindrome(self, head: ListNode) -> bool:
# self.head2 = head
# return self.recursion(head)
if not head or not head.next: return True
# 找出中间节点，奇数点归前半段
slow, fast = head, head
while fast.next and fast.next.next:
slow = slow.next
fast = fast.next.next
# 反转后续节点
rehead = self.reverseList(slow.next)
# 判断回文
flag, head2 = True, rehead
while head2 and flag:
if head.val != head2.val: flag = False
head, head2 = head.next, head2.next
# 反转回去
slow.next = self.reverseList(rehead)
return flag
# @lc code=end
Solution().isPalindrome(createSLink([1, 2, 3, 2, 1]))