Day 3 链表1
之前写c的时候就被链表折磨过很多次，现在准备用python 重温一下。
链表的基础知识
相关阅读：https://zhuanlan.zhihu.com/p/60057180
https://mp.weixin.qq.com/mp/appmsgalbum?__biz=MzI0NjAxMDU5NA==&action=getalbum&album_id=1928300901752963079&scene=173&from_msgid=2475918898&from_itemidx=1&count=3&nolastread=1#wechat_redirect
Leetcode 203:
def removeElements(self, head, val) :
#定义一个辅助节点指向当前头节点
dummy_head = ListNode(next=head)
cur = dummy_head
while cur.next :
if cur.next.val == val:
cur.next = cur.next.next
else:
cur = cur.next
return dummy_head.next Leetcode 707
class ListNode(object) :
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class MyLinkedList(object) :
def __init__(self) :
self.dummy = ListNode()
self.size = 0
def get(self, index):
if index < 0 or index >= self.size:
return -1
cur = self.dummy.next
for i in range(index) :
cur = cur.next
return cur.val
def addAtHead(self, val):
return self.addAtIndex(0, val)
def addAtTail(self, val):
return self.addAtIndex(self.size, val)
def addAtIndex(self, index, val):
if index < 0 or index > self.size:
return
cur = self.dummy
add_node = ListNode(val)
for i in range(index):
cur = cur.next
add_node.next = cur.next
cur.next = add_node
self.size += 1
def deleteAtIndex(self, index):
if index < 0 or index >= self.size :
return
cur = self.dummy
for i in range(index) :
cur = cur.next
cur.next = cur.next.next
self.size -= 1
def items(self):
cur = self.dummy
while cur is not None:
yield cur.val
cur = cur.next
Leetcode 206
def reverseList(self, head) :
cur = head
pre = None
while cur :
temp = cur.next
cur.next = pre
pre = cur
cur = temp
return pre