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21. 合并两个有序链表
将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例:
输入:1->2->4, 1->3->4 输出:1->1->2->3->4->4
代码:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def mergeTwoLists(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
#### the first method
newHead = ListNode(0)
pre = newHead
while l1 and l2:
if l1.val < l2.val:
newHead.next = l1
l1 = l1.next
else:
newHead.next = l2
l2 = l2.next
newHead = newHead.next
if l1:
newHead.next = l1
list.append(pre.next.val)
elif l2:
newHead.next = l2
list.append(pre.next.val)
return pre.next
#### the second method
if l1 == None:
return l2
if l2 == None:
return l1
dummy = ListNode(0)
tmp = dummy
while l1 and l2:
if l1.val <= l2.val:
tmp.next = l1
l1 = l1.next
tmp = tmp.next
else:
tmp.next = l2
l2 = l2.next
tmp = tmp.next
if l2 == None:
tmp.next = l1
else:
tmp.next = l2
return dummy.next
141. 环形链表
给定一个链表,判断链表中是否有环。
代码;
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def hasCycle(self, head):
"""
:type head: ListNode
:rtype: bool
"""
#### the first method
# slow = fast = head
# if head == None or head.next == None:
# return False
# while fast and fast.next:
# slow = slow.next
# fast = fast.next.next
# if fast == slow:
# return True
# return False
#### the second method
slow = fast = head
while fast:
slow, fast = slow.next, fast.next
if not fast:
return False
fast = fast.next
if fast == slow:
return True
return False
234. 回文链表
给定一个链表,判断链表中是否有环。
请判断一个链表是否为回文链表。
示例 1:
输入: 1->2 输出: false
示例 2:
输入: 1->2->2->1 输出: true
代码:
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def isPalindrome(self, head):
"""
:type head: ListNode
:rtype: bool
"""
##### the first method
# if head == None or head.next == None:
# return True
# values = []
# p = head
# while p.next:
# values.append(p.val)
# p = p.next
# values.append(p.val)
# return values[:] == values[::-1]
##### the second method
if head is None or head.next is None:
return True
mid = ListNode(0)
mid.next = head
fast = mid
while fast and fast.next:
mid = mid.next
fast = fast.next.next
pre = mid
cur = mid.next
while cur:
temp = cur.next
cur.next = pre
pre = cur
cur = temp
start = head
end = pre
if start.val != end.val:
return False
while start != mid.next:
if start.val != end.val:
return False
start = start.next
end = end.next
return True
给定一个链表,判断链表中是否有环。