1.反转链表：
class Solution:
# 返回ListNode
def ReverseList(self, pHead):
# write code here
if pHead == None or pHead.next == None:#若链表为空或只有一个数就直接返回0
return pHead
pre = None#定义一个空链表
while pHead:
nex = pHead.next
pHead.next = pre
pre = pHead#指针后移
pHead = nex#指针后移
return pre
2.数组排序 可直接return sorted(arr)
class Solution:
def MySort(self , arr ):
len_1 = len(arr)
list_1 = []
for i in range(len_1):
list_1.append(arr[i])
list_1.sort()
return list_1
3.判断链表是否有环
class Solution:
def hasCycle(self , head ):
# write code here
if head is None or head.next is None:
return False
first = head.next
last = head.next.next
while last is not None:
if first.val == last.val:
return True
first = first.next
if last.next is None or last.next.next is None:
return False
last = last.next.next
return False
4.分别按照二叉树先序，中序和后序打印所有的节点。
class Solution:
def threeOrders(self , root ):
ans = [[],[],[]]
def dfs(root):
if not root:
return
ans[0].append(root.val)
dfs(root.left)
ans[1].append(root.val)
dfs(root.right)
ans[2].append(root.val)
return
dfs(root)
return ans
5.二分查找
class Solution:
def upper_bound_(self , n , v , a ):
# write code here
if a[-1]<v:
return n+1
left = 0
right = n-1
while left<right:
mid = (left+right)//2
if a[mid]>=v:
right = mid
if a[mid]<v:
left = mid+1
return left+1