题目：
第一天：707. 设计链表
第二天：206. 反转链表
收获，在使用head时，要先将self.head赋值给head，才能对head进行操作，在add和delete之后要改变链表的长度，range（）函数是从0开始的
class Node:
def __init__(self, val):
self.val = val
self.next = None
class MyLinkedList:
def __init__(self):
self.head = Node(0)
self.size = 0
def get(self, index: int) -> int:
if index < 0 or index >= self.size:
return -1
head = self.head
#返回的是第i个,range()是从0开始算的
for i in range(index + 1):
head = head.next
return head.val
def addAtHead(self, val: int) -> None:
new = Node(val)
temp = self.head.next
self.head.next = new
new.next = temp
self.size += 1
def addAtTail(self, val: int) -> None:
new = Node(val)
head = self.head
while head.next != None:
head = head.next
head.next = new
self.size += 1
def addAtIndex(self, index: int, val: int) -> None:
'''
if index > self.number: return
elif index = n:
self.number.next = val
else:
self.
'''
new = Node(val)
head = self.head
if index == 0:
self.addAtHead(val)
return
elif index == self.size:
self.addAtTail(val)
return
elif index < 0 or index > self.size:
return
for i in range(index):
head = head.next
temp = head.next
head.next = new
new.next = temp
self.size += 1
def deleteAtIndex(self, index: int) -> None:
'''
if i == index - 1 and 0 < index < len(self):
self.next = self.next.next
elif index < 0:
self.
'''
if index >= self.size:
return
elif index < 0:
index = 0
head = self.head
for i in range(index):
head = head.next
head.next = head.next.next
self.size -= 1
# Your MyLinkedList object will be instantiated and called as such:
# obj = MyLinkedList()
# param_1 = obj.get(index)
# obj.addAtHead(val)
# obj.addAtTail(val)
# obj.addAtIndex(index,val)
# obj.deleteAtIndex(index)
2.在反转的过程中，首先要用tmp保留整个链表的结构，以便遍历。在循环中让cur.next指向pre，使得cur永远是新添加的结点
class Solution:
def reverseList(self, head: ListNode) -> ListNode:
cur = head
pre = None
while (cur != None):
temp = cur.next #将原本的cur.next先存起来
cur.next = pre #反转
pre = cur #让把遍历到的cur链表赋给pre，
cur = temp #遍历下一个结点
return pre