简单列表：包含 val 和next指针
双端链表： 包含val 和 pre、next指针
二、定义链表class ListNode_handle:
def __init__(self):
self.cur_node = None
def add(self, data):
# add a new node pointed to previous node
node = ListNode()
node.val = data
node.next = self.cur_node
self.cur_node = node
return node
def print_ListNode(self, node):
while node:
print('nnode: ', node, ' value: ', node.val, ' next: ', node.next)
node = node.next
def reverse(self, nodelist):
list = []
while nodelist:
list.append(nodelist.val)
nodelist = nodelist.next
result = ListNode()
result_handle = ListNode_handle()
for i in list:
result = result_handle.add(i)
return result
# node_list = [5,3,4,1,6]
list_handle = ListNode_handle()
l1 = ListNode()
l1_list = [1, 8, 3]
for i in l1_list:
l1 = list_handle.add(i)
l1 = list_handle.reverse(l1)
list_handle.print_ListNode(l1)
输出结果：
三、删除链表中的一个元素删除链表中的元素思想就是找到这个元素的下一个元素赋值给需要删除的元素，然后把当前这个元素指针指向需要删除元素的后两个值
class ListNode1:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
"""
@param: head: a ListNode
@param: val: An integer
@return: a ListNode
"""
def removeElements(self, head, val):
# write your code here
tem_head = tem = ListNode1(-1)
head_tem = head
# 遍历一遍链表然后发现和val不相等就加入tem后
while head_tem:
if head_tem.val != val:
tem.next = head_tem
tem = tem.next
# 无论相不相等都需要继续遍历，
head_tem = head_tem.next
# 最后结尾的时候一定要加上None，否则tem的下一位很有可能还有其他元素
tem.next = None
return tem_head.next
head = l1
val = 3
res = Solution().removeElements(head, val)
# list_handle.print_ListNode(l1)
while l1:
print(f'node---: {l1},value:{l1.val},next:{l1.next}')
l1 = l1.next
输出结果：
四、如何合并两个有序链表# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def mergeTwoLists(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
prehead = ListNode(-1)
prev = prehead
while l1 and l2:
if l1.val <= l2.val:
prev.next = l1
l1 = l1.next
else:
prev.next = l2
l2 = l2.next
prev = prev.next
# 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
prev.next = l1 if l1 is not None else l2
return prehead.next
输出结果：
