from ListNode.List_node_Base import ListNode, generateList, printList
def isPalindrome(head):
vals = []
current_node = head
while current_node:
vals.append(current_node.val)
current_node = current_node.next
return vals == vals[::-1]
l1 = generateList([1, 2, 3, 3, 2, 1])
print(isPalindrome(l1))
from ListNode.List_node_Base import ListNode, generateList, printList
class Solution(object):
def mergeKLists(self, lists):
nodes = []
head = point = ListNode(0)
for l in lists:
while l:
nodes.append(l.val)
l = l.next
for x in sorted(nodes):
point.next = ListNode(x)
point = point.next
return head.next
if __name__ == '__main__':
l1 = generateList([1, 4, 5])
l2 = generateList([1, 3, 4])
l3 = generateList([2, 6])
m = Solution()
merage = m.mergeKLists([l1, l2, l3])
printList(merage)
from ListNode.List_node_Base import ListNode, generateList, printList
class Solution:
def mergeTwoLists(self, l1, l2):
if l1 is None:
return l2
elif l2 is None:
return l1
elif l1.val < l2.val:
l1.next = self.mergeTwoLists(l1.next, l2)
return l1
else:
l2.next = self.mergeTwoLists(l1, l2.next)
return l2
l1 = generateList([1, 2, 4])
l2 = generateList([1, 3, 4])
a = Solution()
k = a.mergeTwoLists(l1, l2)
printList(k)
from ListNode.List_node_Base import ListNode, generateList, printList
class Solution:
def reverseList(self, head):
new_head = None
while head:
tmp = head.next # 备份原来head节点的next地址
head.next = new_head
new_head = head
head = tmp
return new_head
l1 = generateList([1, 3, 7])
a = Solution()
k = a.reverseList(l1)
printList(k)
from ListNode.List_node_Base import ListNode, generateList, printList
class Solution:
def sortList(self, head): # 放入list排序,放回ListNode
l = []
p = head
while p:
l.append(p.val)
p = p.next
l.sort()
p = head
for e in l:
p.val = e
p = p.next
return head
l1 = generateList([5, 1, 2, 5, 15, 34])
a = Solution()
k = a.sortList(l1)
printList(k)
while n:
start = start.next
n -= 1
while start.next != None:
start = start.next
end = end.next
print('+++', start.val, end.val)
end.next = end.next.next
print('___', start.val, end.val)
return pre.next
if __name__ == '__main__':
head = generateList([1, 2, 3, 4, 5, 6, 7])
a = Solution()
head2 = a.method1(head, 3)
printList(head2)
'''
方法二: 使用双指针
由于头节点有可能被删除,故设置 pre 指针指向 head
设置快慢指针 前后指针 start, end, 指向 pre. 先让 end 走 n 步, 再将 start 和 end 一起移动.当 end.next 为 None时, 也就是 end 到了尾节点时, start 真好走到待删除节点的前一个节点.
此时 让 start.next = start.next.next, 即可删除 倒数第 n 个节点. 返回 pre.next 即可. 此处不返回 head 是因为 head 可能被删除.
作者：wd6
链接：https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/solution/shan-chu-lian-biao-de-di-n-ge-jie-dian-pythonliang/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
'''
def addTwoNumbers(self, l1, l2):
# 定义一个进位标志
a, b, p, carry = l1, l2, None, 0
while a or b:
# a和b节点的值相加，如果有进位还要加上进位的值
val = (a.val if a else 0) + (b.val if b else 0) + carry
carry = 1 if val >= 10 else 0 # 根据val值判断,不是0,就是1
val = val % 10 # 取val十位上的数
p = a if a else b
p.val = val
# a和b指针都前进一位
a, b = a.next if a else None, b.next if b else None
p.next = a
# 根据a和b是否为空，p指针也前进一位
# 不要忘记最后的边界条件，如果循环结束carry>0说明有进位需要处理这个条件
if carry:
p.next = ListNode(1)
# 每次迭代实际上都是将val赋给a指针的，所以最后返回的是l1
return l1
if __name__ == "__main__":
l1 = generateList([7, 7, 7, 0])
l2 = generateList([6, 6, 6])
printList(l1)
printList(l2)
s = Solution()
sum = s.addTwoNumbers(l1, l2)
printList(sum)